Is it possible to find all integer solutions to this system of equations, inequalities and inequations? Given the following three equations, assuming all unknowns are integers:
$$
x y =\left(x + y - z \right) a  
\\
 x z =\left(x + z - y \right) b  
\\
 y z =\left(y + z - x \right) c  
$$
And the following constraints on the values for $x$, $y$, and $z$:
$$
0 < z \lt x + y \\
0 < y \lt x + z \\
0 < x \lt y + z \\
x \ne y \\
x \ne z \\
y \ne z 
$$
There are no constraints on the values of $a$, $b$, and $c$ beyond that they are integers.
Is there any tractable way to find all of the sets of integer solutions that satisfy the constraints, or lacking that, a way to prove that no integer solutions for all unknowns that satisfy the constraints exist?

Also, let $p=x+y-z$, $q=x+z-y$, $r = y+z-x$ for positive integer $p$, $q$, $r$ (which is possible because of the inequality constraints). Then $2x=p+q$, $2y=p+r$ and $2z=r+q$, so the equations are $(p+q)(p+r)=4ap$, $(p+q)(r+q)=4bq$, $(p+r)(r+q)=4cr$, which means that $p$ is a factor of $qr$, $q$ is a factor of $pr$ and $r$ is a factor of $pq$. This gives another set of solutions: pick $p = 4st$, $q=4tu$, $r=4us$, so that $a=(s+u)(t+u)$, $b=(s+u)(s+t)$, $c=(t+u)(s+t)$ and $x=2t(u+s)$, $y=2s(u+t)$ and $z=2u(s+t)$. For example, $s=1$, $t=2$, $u=3$ gives $x=16$, $y=10$ and $z=18$, with $a =20$, $b= 12$, $c=15$.
