According to your comment, your plane is $P=\{u\alpha+v\beta+b\colon(u,v)\in\mathbb R^2\}$, where $\alpha$ and $\beta$ are two orthogonal vectors in $\mathbb R^n$ and $b\in\mathbb R^n$. Assuming that the vectors $\alpha$ and $\beta$ are nonzero, let $(a_1,\dots,a_n)$ be any basis of $\mathbb R^n$ such that $a_1=\alpha$ and $a_2=\beta$. Let the random variables $Y_1,\dots,Y_n$ be the coordinates of the random vector $X-b$, where $X:=[X_1,\dots,X_n]^T$, so that $X-b=AY$, where $A$ is the $n\times n$ matrix with columns $a_1,\dots,a_n$ and $Y:=[Y_1,\dots,Y_n]^T$.
Then the joint pdf of $X_1,\dots,X_n$ is given by the formula
$$f_X(x)=f(x_1)\cdots f(x_n)\tag{1}$$
for $x=(x_1,\dots,x_n)\in\mathbb R^n$, where $f$ is the pdf of $\text{Logistic}(0,1)$, and the joint pdf of $Y_1,\dots,Y_n$ is given by the formula
$$f_Y(y)=f_X(b+Ay)|\det A| \tag{2}$$
for $y=(y_1,\dots,y_n)\in\mathbb R^n$.
So, the joint pdf of the 2D conditional distribution you want to sample from is given by the formula
$$f_{Y_1,Y_2|Y_3,\dots,Y_n}(y_1,y_2|y_3=\dots=y_n=0)
=\frac{f_Y(y_1,y_2,0,\dots,0)}{f_{Y_3,\dots,Y_n}(0,\dots,0)}
\propto f_Y(y_1,y_2,0,\dots,0)$$
for $(y_1,y_2)\in\mathbb R^2$, where $f_{Y_3,\dots,Y_n}$ is the joint pdf of $Y_3,\dots,Y_n$ and $\propto$ means "proportional to".
So, if you don't want to evaluate the integral
$$f_{Y_3,\dots,Y_n}(0,\dots,0)=\int_{\mathbb R^2}dy_1\,dy_2\, f_Y(y_1,y_2,0,\dots,0), $$
then to sample from the 2D conditional distribution in question you can use one of the Markov chain Monte Carlo (MCMC) methods, which "create samples from a continuous random variable, with probability density proportional to a known function." In your case, the known function is $(y_1,y_2)\mapsto f_Y(y_1,y_2,0,\dots,0)$, with $f_Y$ computed according to (2) and (1).
See e.g. this answer on how to realize MCMC methods in Mathematica.