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$\newcommand\R{\mathbb R}\newcommand\1{\mathbf1}$When you say "I want to integrate over $\mathbb{R}^k$ restricted to the plain where $\sum_{i=1}^{k}y_i=x$", you have to specify the measure over the plane over which you want to integrate.

It appears you want this measure to be induced by the Lebesgue measure on $\R^k$. Then the integration can be done as follows. Let $$c:=N^2r\in(0,\infty)$$ and $$t:=x/\sqrt k,$$ the (signed) distance from the the origin to your plane $$\Pi_t:=\{y\in\R^k\colon u\cdot y=t\}=\{y\in\R^k\colon \1\cdot y=x\},$$ where $\cdot$ denotes the dot product, $\1:=(1,\dots,1)\in\R^k$, and $$u:=\1/\sqrt k$$ is a unit normal vector to the plane $\Pi_t$. Thus, instead of the parameter $x$, we use the more geometrical parameter $t$.

Then the integral in question can be written as $$I_t:=e^{ct^2}J_t,\quad\text{where}\quad J_t:=\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2},$$ $|y|$ is the Euclidean norm of $y$, and, for each real $t$, $\mu_t$ is the measure over the plane $\Pi_t$ induced by the Lebesgue measure on $\R^k$ in the following sense: \begin{equation} \int_a^b dt\, \int_{\Pi_t}\mu_t(dy)g(y) =\int_{\Pi_{a,b}}dy\,g(y) \tag{1} \end{equation} for all nonnegative Borel-measurable functions $g\colon\R^k\to\R$ and all real $a$ and $b$ such that $a<b$, where $$\Pi_{a,b}:=\bigcup_{t\in[a,b]}\Pi_t=\{y\in\R^k\colon a\le u\cdot y\le b\}.$$

Then for such $a$ and $b$ we have $$\int_a^b dt\, J_t=\int_a^b dt\, \int_{\Pi_t}\mu_t(dy)e^{-c|y|^2} =K_{a,b}:=\int_{\Pi_{a,b}}dy\,e^{-c|y|^2}.$$ (See the remark on this at the end of this answer.)

To compute the integral $K_{a,b}$, let us use a substitution of the form $y=Qz$, where $Q$ is any orthogonal $k\times k$ matrix whose first column is the unit vector $u$, so that $y=Qz$ implies $z_1=u\cdot y$, where $z_j$ is the $j$'s coordinate of the vector $z$; such an orthogonal matrix $Q$ exists. Then we can write $$K_{a,b}=\int_{\R^k}dy\,e^{-c|y|^2}\,1(a\le u\cdot y\le b) \\ =\int_{\R^k}dz\,e^{-c|z|^2}\,1(a\le z_1\le b) \\ =\int_a^b dz_1\,e^{-cz_1^2}\int_{\R^{k-1}}dw\,e^{-c|w|^2} \\ =\int_a^b dz_1\,e^{-cz_1^2}\,(\pi/c)^{(k-1)/2}. $$ So, $$J_t=\frac d{dt}\,K_{a,t}=e^{-ct^2}\,(\pi/c)^{(k-1)/2}.$$ Thus, the integral in question is $$I_t:=e^{ct^2}J_t=(\pi/c)^{(k-1)/2}=\Big(\frac\pi{N^2r}\Big)^{(k-1)/2}.$$

This differs from both of your answers -- but you never defined the measure over which you integrate.

Remark: Intuitively, think of $a$ and $b$ as being close to a real number $t$, and hence to each other. We approximate the integral $\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2}$ over the plane $\Pi_t$ by $\frac1{b-a}\,\int_{\Pi_{a,b}}dy\, e^{-c|y|^2}$, that is, by the integral over the thin layer $\Pi_{a,b}$ between two parallel planes $\Pi_a$ and $\Pi_b$ (close to the plane $\Pi_t$) divided by thickness $b-a$ of the layer.

Formally, we are dealing here with disintegration of a measure. That linked theorem deals only with probability measures, but it is trivially extended to finite measures. If we forget about this finiteness condition for a moment, then in that linked theorem we can choose $X=\R$, $Y=\R^k$, let the map $\pi\colon Y\to X$ be the projection map defined by $\pi(y):=u\cdot y$ for all $y\in Y=\R^k$, and let $\mu:=\lambda_k$ and $\nu:=\lambda_1$, where $\lambda_k$ is the Lebesgue measure over $\R^k$. I think the finiteness condition in that linked theorem is inessential, and the proof will hold for any Borel measures $\mu$, at least if $\mu$ is $\sigma$-finite. Alternatively, one can approximate here the Lebesgue measure over $\R^k$ by the finite Lebesgue measures over big cubes in $\R^k$.

$\newcommand\R{\mathbb R}\newcommand\1{\mathbf1}$When you say "I want to integrate over $\mathbb{R}^k$ restricted to the plane where $\sum_{i=1}^{k}y_i=x$", you have to specify the measure over the plane over which you want to integrate.

It appears you want this measure to be induced by the Lebesgue measure on $\R^k$. Then the integration can be done as follows. Let $$c:=N^2r\in(0,\infty)$$ and $$t:=x/\sqrt k,$$ the (signed) distance from the the origin to your plane $$\Pi_t:=\{y\in\R^k\colon u\cdot y=t\}=\{y\in\R^k\colon \1\cdot y=x\},$$ where $\cdot$ denotes the dot product, $\1:=(1,\dots,1)\in\R^k$, and $$u:=\1/\sqrt k$$ is a unit normal vector to the plane $\Pi_t$. Thus, instead of the parameter $x$, we use the more geometrical parameter $t$.

Then the integral in question can be written as $$I_t:=e^{ct^2}J_t,\quad\text{where}\quad J_t:=\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2},$$ $|y|$ is the Euclidean norm of $y$, and, for each real $t$, $\mu_t$ is the measure over the plane $\Pi_t$ induced by the Lebesgue measure on $\R^k$ in the following sense: \begin{equation} \int_a^b dt\, \int_{\Pi_t}\mu_t(dy)g(y) =\int_{\Pi_{a,b}}dy\,g(y) \tag{1} \end{equation} for all nonnegative Borel-measurable functions $g\colon\R^k\to\R$ and all real $a$ and $b$ such that $a<b$, where $$\Pi_{a,b}:=\bigcup_{t\in[a,b]}\Pi_t=\{y\in\R^k\colon a\le u\cdot y\le b\}.$$

Then for such $a$ and $b$ we have $$\int_a^b dt\, J_t=\int_a^b dt\, \int_{\Pi_t}\mu_t(dy)e^{-c|y|^2} =K_{a,b}:=\int_{\Pi_{a,b}}dy\,e^{-c|y|^2}.$$ (See the remark on this at the end of this answer.)

To compute the integral $K_{a,b}$, let us use a substitution of the form $y=Qz$, where $Q$ is any orthogonal $k\times k$ matrix whose first column is the unit vector $u$, so that $y=Qz$ implies $z_1=u\cdot y$, where $z_j$ is the $j$'s coordinate of the vector $z$; such an orthogonal matrix $Q$ exists. Then we can write $$K_{a,b}=\int_{\R^k}dy\,e^{-c|y|^2}\,1(a\le u\cdot y\le b) \\ =\int_{\R^k}dz\,e^{-c|z|^2}\,1(a\le z_1\le b) \\ =\int_a^b dz_1\,e^{-cz_1^2}\int_{\R^{k-1}}dw\,e^{-c|w|^2} \\ =\int_a^b dz_1\,e^{-cz_1^2}\,(\pi/c)^{(k-1)/2}. $$ So, $$J_t=\frac d{dt}\,K_{a,t}=e^{-ct^2}\,(\pi/c)^{(k-1)/2}.$$ Thus, the integral in question is $$I_t:=e^{ct^2}J_t=(\pi/c)^{(k-1)/2}=\Big(\frac\pi{N^2r}\Big)^{(k-1)/2}.$$

This differs from both of your answers -- but you never defined the measure over which you integrate.

Remark: Intuitively, think of $a$ and $b$ as being close to a real number $t$, and hence to each other. We approximate the integral $\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2}$ over the plane $\Pi_t$ by $\frac1{b-a}\,\int_{\Pi_{a,b}}dy\, e^{-c|y|^2}$, that is, by the integral over the thin layer $\Pi_{a,b}$ between two parallel planes $\Pi_a$ and $\Pi_b$ (close to the plane $\Pi_t$) divided by thickness $b-a$ of the layer.

Formally, we are dealing here with disintegration of a measure. That linked theorem deals only with probability measures, but it is trivially extended to finite measures. If we forget about this finiteness condition for a moment, then in that linked theorem we can choose $X=\R$, $Y=\R^k$, let the map $\pi\colon Y\to X$ be the projection map defined by $\pi(y):=u\cdot y$ for all $y\in Y=\R^k$, and let $\mu:=\lambda_k$ and $\nu:=\lambda_1$, where $\lambda_k$ is the Lebesgue measure over $\R^k$. I think the finiteness condition in that linked theorem is inessential, and the proof will hold for any Borel measures $\mu$, at least if $\mu$ is $\sigma$-finite. Alternatively, one can approximate here the Lebesgue measure over $\R^k$ by the finite Lebesgue measures over big cubes in $\R^k$.

$\newcommand\R{\mathbb R}\newcommand\1{\mathbf1}$When you say "I want to integrate over $\mathbb{R}^k$ restricted to the plain where $\sum_{i=1}^{k}y_i=x$", you have to specify the measure over the plane over which you want to integrate.

It appears you want this measure to be induced by the Lebesgue measure on $\R^k$. Then the integration can be done as follows. Let $$c:=N^2r\in(0,\infty)$$ and $$t:=x/\sqrt k,$$ the (signed) distance from your plane $$\Pi_t:=\{y\in\R^k\colon u\cdot y=t\}=\{y\in\R^k\colon \1\cdot y=x\},$$ where $\cdot$ denotes the dot product, $\1:=(1,\dots,1)\in\R^k$, and $$u:=\1/\sqrt k$$ is a unit normal vector to the plane $\Pi_t$. Thus, instead of the parameter $x$, we use the more geometrical parameter $t$.

Then the integral in question can be written as $$I_t:=e^{ct^2}J_t,\quad\text{where}\quad J_t:=\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2},$$ $|y|$ is the Euclidean norm of $y$, and, for each real $t$, $\mu_t$ is the measure over the plane $\Pi_t$ induced by the Lebesgue measure on $\R^k$ in the following sense: \begin{equation} \int_a^b dt\, \int_{\Pi_t}\mu_t(dy)g(y) =\int_{\Pi_{a,b}}dy\,g(y) \tag{1} \end{equation} for all nonnegative Borel-measurable functions $g\colon\R^k\to\R$ and all real $a$ and $b$ such that $a<b$, where $$\Pi_{a,b}:=\bigcup_{t\in[a,b]}\Pi_t=\{y\in\R^k\colon a\le u\cdot y\le b\}.$$

Then for such $a$ and $b$ we have $$\int_a^b dt\, J_t=\int_a^b dt\, \int_{\Pi_t}\mu_t(dy)e^{-c|y|^2} =K_{a,b}:=\int_{\Pi_{a,b}}dy\,e^{-c|y|^2}.$$ (See the remark on this at the end of this answer.)

To compute the integral $K_{a,b}$, let us use a substitution of the form $y=Qz$, where $Q$ is any orthogonal $k\times k$ matrix whose first column is the unit vector $u$, so that $y=Qz$ implies $z_1=u\cdot y$, where $z_j$ is the $j$'s coordinate of the vector $z$; such an orthogonal matrix $Q$ exists. Then we can write $$K_{a,b}=\int_{\R^k}dy\,e^{-c|y|^2}\,1(a\le u\cdot y\le b) \\ =\int_{\R^k}dz\,e^{-c|z|^2}\,1(a\le z_1\le b) \\ =\int_a^b dz_1\,e^{-cz_1^2}\int_{\R^{k-1}}dw\,e^{-c|w|^2} \\ =\int_a^b dz_1\,e^{-cz_1^2}\,(\pi/c)^{(k-1)/2}. $$ So, $$J_t=\frac d{dt}\,K_{a,t}=e^{-ct^2}\,(\pi/c)^{(k-1)/2}.$$ Thus, the integral in question is $$I_t:=e^{ct^2}J_t=(\pi/c)^{(k-1)/2}=\Big(\frac\pi{N^2r}\Big)^{(k-1)/2}.$$

This differs from both of your answers -- but you never defined the measure over which you integrate.

Remark: Intuitively, think of $a$ and $b$ being close to a real number $t$, and hence to each other. We approximate the integral $\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2}$ over the plane $\Pi_t$ by $\frac1{b-a}\,\int_{\Pi_{a,b}}dy e^{-c|y|^2}$, that is, by the integral over the thin layer $\Pi_{a,b}$ between two parallel planes $\Pi_a$ and $\Pi_b$ (close to the plane $\Pi_t$) divided by thickness $b-a$ of the layer.

Formally, we are dealing here with disintegration of a measure. That linked theorem deals only with probability measures, but it is trivially extended to finite measures. If we forget about this finiteness condition for a moment, then in that linked theorem we can choose $X=\R$, $Y=\R^k$, let the map $\pi\colon Y\to X$ be the projection map defined by $\pi(y):=u\cdot y$ for all $y\in Y=\R^k$, and let $\mu:=\lambda_k$ and $\nu:=\lambda_1$, where $\lambda_k$ is the Lebesgue measure over $\R^k$. I think the finiteness condition in that linked theorem is inessential, and the proof will hold for any Borel measures $\mu$, at least if $\mu$ is $\sigma$-finite. Alternatively, one can approximate here the Lebesgue measure over $\R^k$ by the finite Lebesgue measures over big cubes in $\R^k$.

$\newcommand\R{\mathbb R}\newcommand\1{\mathbf1}$When you say "I want to integrate over $\mathbb{R}^k$ restricted to the plain where $\sum_{i=1}^{k}y_i=x$", you have to specify the measure over the plane over which you want to integrate.

It appears you want this measure to be induced by the Lebesgue measure on $\R^k$. Then the integration can be done as follows. Let $$c:=N^2r\in(0,\infty)$$ and $$t:=x/\sqrt k,$$ the (signed) distance from the the origin to your plane $$\Pi_t:=\{y\in\R^k\colon u\cdot y=t\}=\{y\in\R^k\colon \1\cdot y=x\},$$ where $\cdot$ denotes the dot product, $\1:=(1,\dots,1)\in\R^k$, and $$u:=\1/\sqrt k$$ is a unit normal vector to the plane $\Pi_t$. Thus, instead of the parameter $x$, we use the more geometrical parameter $t$.

Then the integral in question can be written as $$I_t:=e^{ct^2}J_t,\quad\text{where}\quad J_t:=\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2},$$ $|y|$ is the Euclidean norm of $y$, and, for each real $t$, $\mu_t$ is the measure over the plane $\Pi_t$ induced by the Lebesgue measure on $\R^k$ in the following sense: \begin{equation} \int_a^b dt\, \int_{\Pi_t}\mu_t(dy)g(y) =\int_{\Pi_{a,b}}dy\,g(y) \tag{1} \end{equation} for all nonnegative Borel-measurable functions $g\colon\R^k\to\R$ and all real $a$ and $b$ such that $a<b$, where $$\Pi_{a,b}:=\bigcup_{t\in[a,b]}\Pi_t=\{y\in\R^k\colon a\le u\cdot y\le b\}.$$

Then for such $a$ and $b$ we have $$\int_a^b dt\, J_t=\int_a^b dt\, \int_{\Pi_t}\mu_t(dy)e^{-c|y|^2} =K_{a,b}:=\int_{\Pi_{a,b}}dy\,e^{-c|y|^2}.$$ (See the remark on this at the end of this answer.)

To compute the integral $K_{a,b}$, let us use a substitution of the form $y=Qz$, where $Q$ is any orthogonal $k\times k$ matrix whose first column is the unit vector $u$, so that $y=Qz$ implies $z_1=u\cdot y$, where $z_j$ is the $j$'s coordinate of the vector $z$; such an orthogonal matrix $Q$ exists. Then we can write $$K_{a,b}=\int_{\R^k}dy\,e^{-c|y|^2}\,1(a\le u\cdot y\le b) \\ =\int_{\R^k}dz\,e^{-c|z|^2}\,1(a\le z_1\le b) \\ =\int_a^b dz_1\,e^{-cz_1^2}\int_{\R^{k-1}}dw\,e^{-c|w|^2} \\ =\int_a^b dz_1\,e^{-cz_1^2}\,(\pi/c)^{(k-1)/2}. $$ So, $$J_t=\frac d{dt}\,K_{a,t}=e^{-ct^2}\,(\pi/c)^{(k-1)/2}.$$ Thus, the integral in question is $$I_t:=e^{ct^2}J_t=(\pi/c)^{(k-1)/2}=\Big(\frac\pi{N^2r}\Big)^{(k-1)/2}.$$

This differs from both of your answers -- but you never defined the measure over which you integrate.

Remark: Intuitively, think of $a$ and $b$ as being close to a real number $t$, and hence to each other. We approximate the integral $\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2}$ over the plane $\Pi_t$ by $\frac1{b-a}\,\int_{\Pi_{a,b}}dy\, e^{-c|y|^2}$, that is, by the integral over the thin layer $\Pi_{a,b}$ between two parallel planes $\Pi_a$ and $\Pi_b$ (close to the plane $\Pi_t$) divided by thickness $b-a$ of the layer.

Formally, we are dealing here with disintegration of a measure. That linked theorem deals only with probability measures, but it is trivially extended to finite measures. If we forget about this finiteness condition for a moment, then in that linked theorem we can choose $X=\R$, $Y=\R^k$, let the map $\pi\colon Y\to X$ be the projection map defined by $\pi(y):=u\cdot y$ for all $y\in Y=\R^k$, and let $\mu:=\lambda_k$ and $\nu:=\lambda_1$, where $\lambda_k$ is the Lebesgue measure over $\R^k$. I think the finiteness condition in that linked theorem is inessential, and the proof will hold for any Borel measures $\mu$, at least if $\mu$ is $\sigma$-finite. Alternatively, one can approximate here the Lebesgue measure over $\R^k$ by the finite Lebesgue measures over big cubes in $\R^k$.

$\newcommand\R{\mathbb R}\newcommand\1{\mathbf1}$When you say "I want to integrate over $\mathbb{R}^k$ restricted to the plain where $\sum_{i=1}^{k}y_i=x$", you have to specify the measure over the plane over which you want to integrate.

It appears you want this measure to be induced by the Lebesgue measure on $\R^k$. Then the integration can be done as follows. Let $$c:=N^2r\in(0,\infty)$$ and $$t:=x/\sqrt k,$$ the (signed) distance from your plane $$\Pi_t:=\{y\in\R^k\colon u\cdot y=t\}=\{y\in\R^k\colon \1\cdot y=x\},$$ where $\cdot$ denotes the dot product, $\1:=(1,\dots,1)\in\R^k$, and $$u:=\1/\sqrt k$$ is a unit normal vector to the plane $\Pi_t$. Thus, instead of the parameter $x$, we use the more geometrical parameter $t$.

Then the integral in question can be written as $$I_t:=e^{ct^2}J_t,\quad\text{where}\quad J_t:=\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2},$$ $|y|$ is the Euclidean norm of $y$, and, for each real $t$, $\mu_t$ is the measure over the plane $\Pi_t$ induced by the Lebesgue measure on $\R^k$ in the following sense: $$\int_a^b dt\, \int_{\Pi_t}\mu_t(dy)g(y) =\int_{\Pi_{a,b}}dy\,g(y)$$ for all nonnegative Borel-measurable functions $g\colon\R^k\to\R$ and all real $a$ and $b$ such that $a<b$, where $$\Pi_{a,b}:=\bigcup_{t\in[a,b]}\Pi_t=\{y\in\R^k\colon a\le u\cdot y\le b\}.$$

Then for such $a$ and $b$ we have $$\int_a^b dt\, J_t=\int_a^b dt\, \int_{\Pi_t}\mu_t(dy)e^{-c|y|^2} =K_{a,b}:=\int_{\Pi_{a,b}}dy\,e^{-c|y|^2}.$$

Remark: Intuitively, think of $a$ and $b$ being close to a real number $t$, and hence to each other. We approximate the integral $\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2}$ over the plane $\Pi_t$ by $\frac1{b-a}\,\int_{\Pi_{a,b}}dy e^{-c|y|^2}$, that is, by the integral over the thin layer $\Pi_{a,b}$ between two parallel planes $\Pi_a$ and $\Pi_b$ (close to the plane $\Pi_t$) divided by thickness $b-a$ of the layer.

To compute the integral $K_{a,b}$, let us use a substitution of the form $y=Qz$, where $Q$ is any orthogonal $k\times k$ matrix whose first column is the unit vector $u$, so that $y=Qz$ implies $z_1=u\cdot y$, where $z_j$ is the $j$'s coordinate of the vector $z$; such an orthogonal matrix $Q$ exists. Then we can write $$K_{a,b}=\int_{\R^k}dy\,e^{-c|y|^2}\,1(a\le u\cdot y\le b) \\ =\int_{\R^k}dz\,e^{-c|z|^2}\,1(a\le z_1\le b) \\ =\int_a^b dz_1\,e^{-cz_1^2}\int_{\R^{k-1}}dw\,e^{-c|w|^2} \\ =\int_a^b dz_1\,e^{-cz_1^2}\,(\pi/c)^{(k-1)/2}. $$ So, $$J_t=\frac d{dt}\,K_{a,t}=e^{-ct^2}\,(\pi/c)^{(k-1)/2}.$$ Thus, the integral in question is $$I_t:=e^{ct^2}J_t=(\pi/c)^{(k-1)/2}=\Big(\frac\pi{N^2r}\Big)^{(k-1)/2}.$$

This differs from both of your answers -- but you never defined the measure over which you integrate.

$\newcommand\R{\mathbb R}\newcommand\1{\mathbf1}$When you say "I want to integrate over $\mathbb{R}^k$ restricted to the plain where $\sum_{i=1}^{k}y_i=x$", you have to specify the measure over the plane over which you want to integrate.

It appears you want this measure to be induced by the Lebesgue measure on $\R^k$. Then the integration can be done as follows. Let $$c:=N^2r\in(0,\infty)$$ and $$t:=x/\sqrt k,$$ the (signed) distance from your plane $$\Pi_t:=\{y\in\R^k\colon u\cdot y=t\}=\{y\in\R^k\colon \1\cdot y=x\},$$ where $\cdot$ denotes the dot product, $\1:=(1,\dots,1)\in\R^k$, and $$u:=\1/\sqrt k$$ is a unit normal vector to the plane $\Pi_t$. Thus, instead of the parameter $x$, we use the more geometrical parameter $t$.

Then the integral in question can be written as $$I_t:=e^{ct^2}J_t,\quad\text{where}\quad J_t:=\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2},$$ $|y|$ is the Euclidean norm of $y$, and, for each real $t$, $\mu_t$ is the measure over the plane $\Pi_t$ induced by the Lebesgue measure on $\R^k$ in the following sense: \begin{equation} \int_a^b dt\, \int_{\Pi_t}\mu_t(dy)g(y) =\int_{\Pi_{a,b}}dy\,g(y) \tag{1} \end{equation} for all nonnegative Borel-measurable functions $g\colon\R^k\to\R$ and all real $a$ and $b$ such that $a<b$, where $$\Pi_{a,b}:=\bigcup_{t\in[a,b]}\Pi_t=\{y\in\R^k\colon a\le u\cdot y\le b\}.$$

Then for such $a$ and $b$ we have $$\int_a^b dt\, J_t=\int_a^b dt\, \int_{\Pi_t}\mu_t(dy)e^{-c|y|^2} =K_{a,b}:=\int_{\Pi_{a,b}}dy\,e^{-c|y|^2}.$$  (See the remark on this at the end of this answer.)

To compute the integral $K_{a,b}$, let us use a substitution of the form $y=Qz$, where $Q$ is any orthogonal $k\times k$ matrix whose first column is the unit vector $u$, so that $y=Qz$ implies $z_1=u\cdot y$, where $z_j$ is the $j$'s coordinate of the vector $z$; such an orthogonal matrix $Q$ exists. Then we can write $$K_{a,b}=\int_{\R^k}dy\,e^{-c|y|^2}\,1(a\le u\cdot y\le b) \\ =\int_{\R^k}dz\,e^{-c|z|^2}\,1(a\le z_1\le b) \\ =\int_a^b dz_1\,e^{-cz_1^2}\int_{\R^{k-1}}dw\,e^{-c|w|^2} \\ =\int_a^b dz_1\,e^{-cz_1^2}\,(\pi/c)^{(k-1)/2}. $$ So, $$J_t=\frac d{dt}\,K_{a,t}=e^{-ct^2}\,(\pi/c)^{(k-1)/2}.$$ Thus, the integral in question is $$I_t:=e^{ct^2}J_t=(\pi/c)^{(k-1)/2}=\Big(\frac\pi{N^2r}\Big)^{(k-1)/2}.$$

This differs from both of your answers -- but you never defined the measure over which you integrate.

Remark: Intuitively, think of $a$ and $b$ being close to a real number $t$, and hence to each other. We approximate the integral $\int_{\Pi_t}\mu_t(dy)e^{-c|y|^2}$ over the plane $\Pi_t$ by $\frac1{b-a}\,\int_{\Pi_{a,b}}dy e^{-c|y|^2}$, that is, by the integral over the thin layer $\Pi_{a,b}$ between two parallel planes $\Pi_a$ and $\Pi_b$ (close to the plane $\Pi_t$) divided by thickness $b-a$ of the layer.

Formally, we are dealing here with disintegration of a measure. That linked theorem deals only with probability measures, but it is trivially extended to finite measures. If we forget about this finiteness condition for a moment, then in that linked theorem we can choose $X=\R$, $Y=\R^k$, let the map $\pi\colon Y\to X$ be the projection map defined by $\pi(y):=u\cdot y$ for all $y\in Y=\R^k$, and let $\mu:=\lambda_k$ and $\nu:=\lambda_1$, where $\lambda_k$ is the Lebesgue measure over $\R^k$. I think the finiteness condition in that linked theorem is inessential, and the proof will hold for any Borel measures $\mu$, at least if $\mu$ is $\sigma$-finite. Alternatively, one can approximate here the Lebesgue measure over $\R^k$ by the finite Lebesgue measures over big cubes in $\R^k$.