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Let me try to reinstate honor to the solution that proposed the basis change, by reducing the general case to the independent ("generic") case via a basis change, as opposed to proving it from scratch. This time I am treating the continuous version. (as in Peter Humphries' solution)

Let $\vec\theta=(\theta_1,\dots,\theta_k)$. Let $S$ be the subspace of vectors $\vec x\in\mathbb R^k$ such that $$\langle \vec a,\vec x\rangle=0$$ for all rational vectors $\vec a\in\mathbb Q^k$ for which $$\langle \vec a,\vec\theta\rangle=0$$ The integer points in $S$ form a lattice $L$ (a discrete subgroup of $\mathbb R^n$), which can be written as $L=\{\,g_1\vec b_1+\dots+g_r\vec b_r\mid g_i\in\mathbb Z\,\}$ for some generating basis vectors $\vec b_1,\dots,\vec b_r\in\mathbb{Z}^k$. Since $S$ is defined by rational equations, this basis spans $S$.
Let $(\theta'_1,\dots,\theta_r')$ be the coordinates of the point $\vec\theta\in S$ with respect to this basis: $$\vec\theta=\theta_1'\vec b_1+\dots+\theta_r'\vec b_r$$ Then $(\theta'_1,\dots,\theta_r')$ is independent over the rationals (see below (*) for a proof). Thus, we can apply the generic continuous Kronecker-Weyl Theorem, and $(t(\theta_1',\dots,\theta_r'))_{t\in \mathbb{R}}$ is uniformly distributed modulo 1 in the $r$-torus $[0,1)^r$. Transforming back to the original coordinates, this means that $(t\vec \theta)_{t\in \mathbb{R}}$ is uniformly distributed modulo $L$ in the fundamental region $$ F = \{\,\lambda_1\vec b_1+\dots+\lambda_r\vec b_r \mid 0\le\lambda_i<1\,\}$$ of the lattice.
Now we map $F$ back into the standard torus $[0,1)^k$ by taking all coordinates modulo 1. No two points of $F$ are mapped to the same point (otherwise we would have a nonzero integer point inside $F$), but "opposite" boundary points are mapped to the same point because they differ by a basis vector $b_i\in\mathbb{Z}^k$. So $F$ forms a nice $r$-dimensional subtorus of $[0,1)^k$.

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(*) Here is the proof that $(\theta'_1,\dots,\theta_r')$ is independent over the rationals. It is not so obvious as I thought. Consider a rational relation $c_1\theta_1'+\dots+c_r\theta'_r=0.$ We can choose inside $S$ a rational vector $\vec a$ such that $\langle \vec a,\vec b_i\rangle = c_i$ for $i=1,\dots,r$. (The vector $\vec a$ is uniquely determined by these equations.) Then $$\langle \vec a, \vec\theta\rangle= \langle \vec a, (\theta_1'\vec b_1+\dots+\theta_r'\vec b_r)\rangle =c_1\theta_1'+\dots+a_r\theta'_r=0 $$ Thus, by the definition of $S$, $\vec a$ should be orthogonal to $S$. Therefore, $\vec a=\vec 0$, and all $c_i$ are $0$.

Let me try to reinstate honor to the solution that proposed the basis change, by reducing the general case to the independent ("generic") case via a basis change, as opposed to proving it from scratch. This time I am treating the continuous version. (as in Peter Humphries' solution)

Let $\vec\theta=(\theta_1,\dots,\theta_k)$. Let $S$ be the subspace of vectors $\vec x\in\mathbb R^k$ such that $$\langle \vec a,\vec x\rangle=0$$ for all rational vectors $\vec a\in\mathbb Q^k$ for which $$\langle \vec a,\vec\theta\rangle=0$$ The integer points in $S$ form a lattice $L$ (a discrete subgroup of $\mathbb R^n$), which can be written as $L=\{\,g_1\vec b_1+\dots+g_r\vec b_r\mid g_i\in\mathbb Z\,\}$ for some generating basis vectors $\vec b_1,\dots,\vec b_r\in\mathbb{Z}^k$. Since $S$ is defined by rational equations, this basis spans $S$.
Let $(\theta'_1,\dots,\theta_r')$ be the coordinates of the point $\vec\theta\in S$ with respect to this basis: $$\vec\theta=\theta_1'\vec b_1+\dots+\theta_r'\vec b_r$$ Then $(\theta'_1,\dots,\theta_r')$ is independent over the rationals (see below (*) for a proof). Thus, we can apply the generic continuous Kronecker-Weyl Theorem, and $(t(\theta_1',\dots,\theta_r'))_{t\in \mathbb{R}}$ is uniformly distributed modulo 1 in the $r$-torus $[0,1)^r$. Transforming back to the original coordinates, this means that $(t\vec \theta)_{t\in \mathbb{R}}$ is uniformly distributed modulo $L$ in the fundamental region $$ F = \{\,\lambda_1\vec b_1+\dots+\lambda_r\vec b_r \mid 0\le\lambda_i<1\,\}$$ of the lattice.
Now we map $F$ back into the standard torus $[0,1)^k$ by taking all coordinates modulo 1. No two points of $F$ are mapped to the same point (otherwise we would have a nonzero integer point inside $F$), but "opposite" boundary points are mapped to the same point because they differ by a basis vector $b_i\in\mathbb{Z}^k$. So $F$ forms a nice $r$-dimensional subtorus of $[0,1)^k$.

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(*) Here is the proof that $(\theta'_1,\dots,\theta_r')$ is independent over the rationals. It is not so obvious as I thought. Consider a rational relation $c_1\theta_1'+\dots+c_r\theta'_r=0.$ We can choose inside $S$ a rational vector $\vec a$ such that $\langle \vec a,\vec b_i\rangle = c_i$ for $i=1,\dots,r$. (The vector $\vec a$ is uniquely determined by these equations.) Then $$\langle \vec a, \vec\theta\rangle= \langle \vec a, (\theta_1'\vec b_1+\dots+\theta_r'\vec b_r)\rangle =c_1\theta_1'+\dots+c_r\theta'_r=0 $$ Thus, by the definition of $S$, $\vec a$ should be orthogonal to $S$. Therefore, $\vec a=\vec 0$, and all $c_i$ are $0$.

Let me try to reinstate honor to the solution that proposed the basis change, by reducing the general case to the independent ("generic") case via a basis change, as opposed to proving it from scratch. This time I am treating the continuous version. (as in Peter Humphries' solution)

Let $\vec\theta=(\theta_1,\dots,\theta_k)$. Let $S$ be the subspace of vectors $\vec x\in\mathbb R^k$ such that $$\langle \vec a,\vec x\rangle=0$$ for all rational vectors $\vec a\in\mathbb Q^k$ for which $$\langle \vec a,\vec\theta\rangle=0$$ The integer points in $S$ form a lattice $L$ (a discrete subgroup of $\mathbb R^n$), which can be written as $L=\{\,g_1\vec b_1+\dots+g_r\vec b_r\mid g_i\in\mathbb Z\,\}$ for some generating basis vectors $\vec b_1,\dots,\vec b_r\in\mathbb{Z}^k$. Since $S$ is defined by rational equations, this basis spans $S$. Let $(\theta'_1,\dots,\theta_r')$ be the coordinates of the point $\vec\theta\in S$ with respect to this basis: $$\vec\theta=\theta_1'\vec b_1+\dots+\theta_r'\vec b_r$$ Then $(\theta'_1,\dots,\theta_r')$ is independent over the rationals (see below (*) for a proof). Thus, we can apply the generic continuous Kronecker-Weyl Theorem, and $(t(\theta_1',\dots,\theta_r'))_{t\in \mathbb{R}}$ is uniformly distributed modulo 1 in the $r$-torus $[0,1)^r$. Transforming back to the original coordinates, this means that $(t\vec \theta)_{t\in \mathbb{R}}$ is uniformly distributed modulo $L$ in the fundamental region $$ F = \{\,\lambda_1\vec b_1+\dots+\lambda_r\vec b_r \mid 0\le\lambda_i<1\,\}$$ of the lattice. Now we map $F$ back into the standard torus $[0,1)^k$ by taking all coordinates modulo 1. No two points of $F$ are mapped to the same point (otherwise we would have a nonzero integer point inside $F$), but "opposite" boundary points are mapped to the same point because they differ by a basis vector $b_i\in\mathbb{Z}^k$. So $F$ forms a nice $r$-dimensional subtorus of $[0,1)^k$.

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(*) Here is the proof that $(\theta'_1,\dots,\theta_r')$ is independent over the rationals. It is not so obvious as I thought. Consider a rational relation $c_1\theta_1'+\dots+c_r\theta'_r=0.$ We can choose inside $S$ a rational vector $\vec a$ such that $\langle \vec a,\vec b_i\rangle = c_i$ for $i=1,\dots,r$. (The vector $\vec a$ is uniquely determined by these equations.) Then $$\langle \vec a, \vec\theta\rangle= \langle \vec a, (\theta_1'\vec b_1+\dots+\theta_r'\vec b_r)\rangle =c_1\theta_1'+\dots+a_r\theta'_r=0 $$ Thus, by the definition of $S$, $\vec a$ should be orthogonal to $S$. Thus, $\vec a=\vec 0$, and all $c_i$ are $0$.

Let me try to reinstate honor to the solution that proposed the basis change, by reducing the general case to the independent ("generic") case via a basis change, as opposed to proving it from scratch. This time I am treating the continuous version. (as in Peter Humphries' solution)

Let $\vec\theta=(\theta_1,\dots,\theta_k)$. Let $S$ be the subspace of vectors $\vec x\in\mathbb R^k$ such that $$\langle \vec a,\vec x\rangle=0$$ for all rational vectors $\vec a\in\mathbb Q^k$ for which $$\langle \vec a,\vec\theta\rangle=0$$ The integer points in $S$ form a lattice $L$ (a discrete subgroup of $\mathbb R^n$), which can be written as $L=\{\,g_1\vec b_1+\dots+g_r\vec b_r\mid g_i\in\mathbb Z\,\}$ for some generating basis vectors $\vec b_1,\dots,\vec b_r\in\mathbb{Z}^k$. Since $S$ is defined by rational equations, this basis spans $S$.
Let $(\theta'_1,\dots,\theta_r')$ be the coordinates of the point $\vec\theta\in S$ with respect to this basis: $$\vec\theta=\theta_1'\vec b_1+\dots+\theta_r'\vec b_r$$ Then $(\theta'_1,\dots,\theta_r')$ is independent over the rationals (see below (*) for a proof). Thus, we can apply the generic continuous Kronecker-Weyl Theorem, and $(t(\theta_1',\dots,\theta_r'))_{t\in \mathbb{R}}$ is uniformly distributed modulo 1 in the $r$-torus $[0,1)^r$. Transforming back to the original coordinates, this means that $(t\vec \theta)_{t\in \mathbb{R}}$ is uniformly distributed modulo $L$ in the fundamental region $$ F = \{\,\lambda_1\vec b_1+\dots+\lambda_r\vec b_r \mid 0\le\lambda_i<1\,\}$$ of the lattice.
Now we map $F$ back into the standard torus $[0,1)^k$ by taking all coordinates modulo 1. No two points of $F$ are mapped to the same point (otherwise we would have a nonzero integer point inside $F$), but "opposite" boundary points are mapped to the same point because they differ by a basis vector $b_i\in\mathbb{Z}^k$. So $F$ forms a nice $r$-dimensional subtorus of $[0,1)^k$.

---

(*) Here is the proof that $(\theta'_1,\dots,\theta_r')$ is independent over the rationals. It is not so obvious as I thought. Consider a rational relation $c_1\theta_1'+\dots+c_r\theta'_r=0.$ We can choose inside $S$ a rational vector $\vec a$ such that $\langle \vec a,\vec b_i\rangle = c_i$ for $i=1,\dots,r$. (The vector $\vec a$ is uniquely determined by these equations.) Then $$\langle \vec a, \vec\theta\rangle= \langle \vec a, (\theta_1'\vec b_1+\dots+\theta_r'\vec b_r)\rangle =c_1\theta_1'+\dots+a_r\theta'_r=0 $$ Thus, by the definition of $S$, $\vec a$ should be orthogonal to $S$. Therefore, $\vec a=\vec 0$, and all $c_i$ are $0$.

Let me try to reinstate honor to the solution that proposed the basis change, by reducing the general case to the independent ("generic") case via a basis change, as opposed to proving it from scratch. This time I am treating the continuous version. (as in Peter Humphries' solution)

Let $\vec\theta=(\theta_1,\dots,\theta_k)$. Let $S$ be the subspace of vectors $\vec x\in\mathbb R^k$ such that $$\langle \vec a,\vec x\rangle=0$$ for all rational vectors $\vec a\in\mathbb Q^k$ for which $$\langle \vec a,\vec\theta\rangle=0$$ The integer points in $S$ form a lattice $L$ (a discrete subgroup of $\mathbb R^n$), which can be written as $L=\{\,g_1\vec b_1+\dots+g_r\vec b_r\mid g_i\in\mathbb Z\,\}$ for some generating basis vectors $\vec b_1,\dots,\vec b_r\in\mathbb{Z}^k$. Since $S$ is defined by rational equations, this basis spans $S$. Let $(\theta'_1,\dots,\theta_r')$ be the coordinates of the point $(\theta_1,\dots,\theta_k)\in S$ with respect to this basis: $$(\theta_1,\dots,\theta_k)=\theta_1'\vec b_1+\dots+\theta_r'\vec b_r$$ Then $(\theta'_1,\dots,\theta_r')$ is independent over the rationals (see below (*) for a proof). Thus, we can apply the generic continuous Kronecker-Weyl Theorem, and $(t(\theta_1',\dots,\theta_r'))_{t\in \mathbb{R}}$ is uniformly distributed modulo 1 in the $r$-torus $[0,1)^r$. Transforming back to the original coordinates, this means that $(t\vec \theta)_{t\in \mathbb{R}}$ is uniformly distributed modulo $L$ in the fundamental region $$ F = \{\,\lambda_1\vec b_1+\dots+\lambda_r\vec b_r \mid 0\le\lambda_i<1\,\}$$ of the lattice. Now we map $F$ back into the standard torus $[0,1)^k$ by taking all coordinates modulo 1. No two points of $F$ are mapped to the same point (otherwise we would have a nonzero integer point inside $F$), but "opposite" boundary points are mapped to the same point because they differ by a basis vector $b_i\in\mathbb{Z}^k$. So $F$ forms a nice $r$-dimensional subtorus of $[0,1)^k$.

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(*) Here is the proof that $(\theta'_1,\dots,\theta_r')$ is independent over the rationals. It is not so obvious as I thought. Suppose there is a nontrivial rational relation $c_1\theta_1'+\dots+c_r\theta'_r=0.$ We can choose inside $S$ a rational vector $\vec a$ such that $\langle \vec a,\vec b_i\rangle = c_i$ for $i=1,\dots,r$. (The vector $\vec a$ is uniquely determined by these equations.) Then $$\langle \vec a, \vec\theta\rangle= \langle \vec a, (\theta_1'\vec b_1+\dots+\theta_r'\vec b_r)\rangle =c_1\theta_1'+\dots+a_r\theta'_r=0 $$ Thus, by the definition of $S$, $a$ should be orthogonal to $S$. This is a contradiction.

Let me try to reinstate honor to the solution that proposed the basis change, by reducing the general case to the independent ("generic") case via a basis change, as opposed to proving it from scratch. This time I am treating the continuous version. (as in Peter Humphries' solution)

Let $\vec\theta=(\theta_1,\dots,\theta_k)$. Let $S$ be the subspace of vectors $\vec x\in\mathbb R^k$ such that $$\langle \vec a,\vec x\rangle=0$$ for all rational vectors $\vec a\in\mathbb Q^k$ for which $$\langle \vec a,\vec\theta\rangle=0$$ The integer points in $S$ form a lattice $L$ (a discrete subgroup of $\mathbb R^n$), which can be written as $L=\{\,g_1\vec b_1+\dots+g_r\vec b_r\mid g_i\in\mathbb Z\,\}$ for some generating basis vectors $\vec b_1,\dots,\vec b_r\in\mathbb{Z}^k$. Since $S$ is defined by rational equations, this basis spans $S$. Let $(\theta'_1,\dots,\theta_r')$ be the coordinates of the point $\vec\theta\in S$ with respect to this basis: $$\vec\theta=\theta_1'\vec b_1+\dots+\theta_r'\vec b_r$$ Then $(\theta'_1,\dots,\theta_r')$ is independent over the rationals (see below (*) for a proof). Thus, we can apply the generic continuous Kronecker-Weyl Theorem, and $(t(\theta_1',\dots,\theta_r'))_{t\in \mathbb{R}}$ is uniformly distributed modulo 1 in the $r$-torus $[0,1)^r$. Transforming back to the original coordinates, this means that $(t\vec \theta)_{t\in \mathbb{R}}$ is uniformly distributed modulo $L$ in the fundamental region $$ F = \{\,\lambda_1\vec b_1+\dots+\lambda_r\vec b_r \mid 0\le\lambda_i<1\,\}$$ of the lattice. Now we map $F$ back into the standard torus $[0,1)^k$ by taking all coordinates modulo 1. No two points of $F$ are mapped to the same point (otherwise we would have a nonzero integer point inside $F$), but "opposite" boundary points are mapped to the same point because they differ by a basis vector $b_i\in\mathbb{Z}^k$. So $F$ forms a nice $r$-dimensional subtorus of $[0,1)^k$.

---

(*) Here is the proof that $(\theta'_1,\dots,\theta_r')$ is independent over the rationals. It is not so obvious as I thought. Consider a rational relation $c_1\theta_1'+\dots+c_r\theta'_r=0.$ We can choose inside $S$ a rational vector $\vec a$ such that $\langle \vec a,\vec b_i\rangle = c_i$ for $i=1,\dots,r$. (The vector $\vec a$ is uniquely determined by these equations.) Then $$\langle \vec a, \vec\theta\rangle= \langle \vec a, (\theta_1'\vec b_1+\dots+\theta_r'\vec b_r)\rangle =c_1\theta_1'+\dots+a_r\theta'_r=0 $$ Thus, by the definition of $S$, $\vec a$ should be orthogonal to $S$. Thus, $\vec a=\vec 0$, and all $c_i$ are $0$.