Integer solutions to $2x - 3y = z$ and $2y - 3z = x$ 
Are there infinitely many integer solutions ($x, y, z$) to the pair of equations $2x-3y=z$ and $2y-3z=x$?

For instance, $(1, 1, 1)$, $(11, 7, 1)$, and $(-11, -7,  -1)$ are solutions. Any help on how to find solutions in general? Thanks!
 A: Note that $(1,1,1)$ does not appear to be a valid solution.
$\begin{cases} 2x-3y=z\\ 2y-3z=x \end{cases}\iff 
\begin{cases} 2(2y-3z)-3y=z\\ x=2y-3z \end{cases}\iff
\begin{cases} y=7z\\ x=14z-3z=11z \end{cases}$
So the solutions are $(11,7,1)\times k$ with $k\in\mathbb Z$.
A: Hint $ $ By Cramer's rule, since the determinant $= 1, $ we have
$$\begin{align} &\begin{bmatrix} 1 &  -2 \\\\  2 &  -3 \end{bmatrix} \begin{bmatrix} x \\\\  y \end{bmatrix} = \begin{bmatrix} -3z \\\\  z\end{bmatrix}\\
\\
\iff\ \begin{bmatrix} x \\\\  y\end{bmatrix} =\, &\begin{bmatrix} -3  &  2 \\\\  -2 &  1 \end{bmatrix} \begin{bmatrix} -3z \\\\  z \end{bmatrix} = \begin{bmatrix} 11z \\\\   7z\end{bmatrix} \end{align}$$
A: $2x - 3y = z$ and $2y - 3z = x$ means  
$2y - 3(2x-3y)= x$
$11y - 7x = 0$
So $y = \frac {7x}{11}$.  So if $x$ is an integer it must be a multiple of $11$.  And if $x = 11a$ then $y = 7a$.
$z = 2x - 3(\frac {7x}{11}) = \frac {x}{11}$.
So any $(11a, 7a, a)$ for $a \in \mathbb Z$ will be a solution.
$(1,1,1)$ is not a solution as $2(1) - 3(1) \ne 1$.
