I have the following affine transformation in 2 dimensions $$ \pmatrix{v\\e}\mapsto\pmatrix{av+be+c\\xv+ye+z} $$ and constraints given as $3v=2e$ and therefore $3(av+be+z)=2(xv+ye+z)$. All variables and parameters are elements of $\mathbb N_0$. I can rewrite this as: $\underbrace{(3a-2x)}_3v+\underbrace{(3c-2z)}_0=\underbrace{(2y-3b)}_2e$. Written using matrices this gets: $$ (3,-2)\cdot \left( \pmatrix{c\\z} + \pmatrix{a&b\\x&y} \pmatrix{v\\e} \right)=0 $$
How to (graphically?) describe the set of solution when $v$ and $e$ are fixed, i.e. in terms of $a,b,c,x,y,z$? Obviously there is an orthogonality involved.
The question is motivated by polyhedra, so the example is the chamfer operation $$ (3,-2)\cdot \left( \pmatrix{0\\0} + \pmatrix{1&2\\0&4} \pmatrix{v\\e} \right)=3v+6e-8e=0\rightarrow3v=2e $$
Setting $c=z=0$ the question boils down to $"$Which matrices fulfill $$ (3,-2)\cdot \pmatrix{a&b\\x&y} \pmatrix{2\\3}=0\\ 6a+9b-4x-6y=0?\text{"} $$
Looks like $ \pmatrix{2\\3}$ should be an eigenvector of $\pmatrix{a&b\\x&y}$ in this case...
Let's look at this from a vector algebra viewpoint. We have $$\mathbf{R} = \left ( \begin{matrix} a & b \\ x & y \end{matrix} \right ) \tag{1}\label{1}$$ as the transformation matrix. Logically, $$\vec{e}_1 = ( a ,\, x ) \tag{2}\label{2}$$ and $$\vec{e}_2 = ( b ,\, y ) \tag{3}\label{3}$$ define the axes in the new coordinate system. We also have $\vec{p} = ( v , e )$ with $3 v = 2 e$, i.e. $e = 3/2 v$. If we use $w = v/2$, then $$\vec{p} = ( 2 w ,\, 3 w ) \tag{4}\label{4}$$ and $$\vec{q} = ( c ,\, z ) \tag{5}\label{5}$$ and finally $$\vec{r} = ( 3 ,\, -2 ) \tag{6}\label{6}$$
All scalars above are nonnegative integers, i.e. $a , b , c , e , v , w , x , y , z \in \mathbb{N}_0$.
Using equations $\eqref{1}$, $\eqref{4}$, $\eqref{5}$, and $\eqref{6}$ we have OP's equation we wish to describe: $$\vec{r} \cdot \left ( \vec{q} + \mathbf{R} \vec{p} \right ) = 0 \tag{7}\label{7}$$
Note that $\eqref{7}$ is equivalent to $$(6 a + 9 b - 4 x - 6 y) w + 3 c - 2 z = 0 \tag{8}\label{8}$$ which yields a pretty straightforward description of the relationship between a constant $w$ and variables $a$, $b$, $c$, $x$, $y$, and $z$.
Note that this also shows that $w = 0$ (i.e. $v = e = 0$) is a special case: If $w = 0$, then the equation simplifies to $3 c = 2 z$, and the values of $a$, $b$, $x$, and $y$ are irrelevant, i.e. matrix $\mathbf{R}$ becomes irrelevant.
Using vector algebra, if we use $\vec{r}_1 = (a , b)$, $\vec{r}_2 = (x , y)$, and $\vec{q} = (c , z)$, then we have $$\vec{p} \cdot \left( 3 \vec{r}_1 - 2 \vec{r}_2 \right ) + \vec{q} \cdot \vec{r} = 0$$ or, moving the dot product to the other side, $$\vec{p} \cdot \left( 2 \vec{r}_2 - 3 \vec{r}_1 \right ) = \vec{q} \cdot \vec{r} \tag{9}\label{9}$$
This means that the projection of $\vec{p}$ to vector $(2\vec{r}_2 - 3\vec{r}_1)$ matches $\vec{q}\cdot\vec{r}$:
where
$$\begin{array}{rcl}
\vec{p} &=& ( 2 w , 3 w ) = ( v , e ) \\
\vec{q} \cdot \vec{r} &=& 3 c - 2 z \\
2\vec{r}_2 - 3\vec{r}_1 &=& ( 2 x - 3 a , 2 y - 3 b )
\end{array}$$
Note that because $\vec{r}$ has a negative second component, $(2\vec{r}_2 - 3\vec{r}_1)$ must have integer ($\mathbb{Z}$) coordinates; they can be negative.
I find this interpretation very interesting, because it basically simplifies the transformation matrix $\mathbf{R}$ into a simple linear combination of its row vectors $\vec{r}_1$ and $\vec{r}_2$.
If we use $$\vec{q} = \vec{p}_0 - \mathbf{R}^{-1} \vec{p}_0$$ we can write equation $\eqref{7}$ as $$\vec{r} \cdot \left ( \vec{p}_0 + \mathbf{R} ( \vec{p} - \vec{p}_0 ) \right ) = 0 \tag{10}\label{10}$$ which can then be described as:
Vector $\vec{r}$ must be orthogonal to vector $\vec{p}$ transformed by $\mathbf{R}$ around point $\vec{p}_0$.
(But, note that $\vec{p}_0$ does not need to have integer or positive coordinates.)
I'm not sure this is an useful description, but it is interesting because vectors $\vec{r}$, $\vec{p}$, and matrix $\mathbf{R}$ vectors $\vec{e}_1$ and $\vec{e}_2$, are all in the positive quadrant of an integer lattice.
We can solve that for any one of the variables. For example, $$a = y - \frac{3}{2} b + \frac{2}{3} x \tag{11}\label{11}$$ which tells us that since $a, b, x, y \in \mathbb{N}_0$, $2 x / 3 + 3 b / 2$ must be a nonnegative integer not greater than $y$. That means $b$ must be even, and $x$ a a multiple of three.
So, if we change to different variables $$\begin{cases}i = x / 3\\ j = y \\ k = b / 2\end{cases}$$ then we can write all suitable matrices as $$\mathbf{R} = \left ( \begin{matrix} 2 i + j - 3 k & 2 k \\ 3 i & j \end{matrix} \right ), \qquad \begin{cases} i, j, k \in \mathbb{N}_0 \\ i + j \gt 0 \\ 3 k \lt 2 i + j \end{cases}$$ where the requirement $i + j \gt 0$ ensures that the second row vector is not a null vector, and the $\lt$ ensures that the first row vector is not a null vector. (I am not completely certain whether degenerate $\mathbf{R}$ should be excluded, but I am assuming so.)