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Remark

Happy Problem-Solving!

We consider the cases $m = -2$ and $m = 1$ first. If $m = 1$ then we obviously need $a = b = c$, and in this case, there are infinitely many solutions. If $m = -2$ then we need $a + b + c = 0$ (just add the inequalities to see this), and in this case there are infinitely many solutions as well (since we get two equations in three variables, and if $(x,y,z)$ is a solution, then so is $(x + k,y + k,z + k)$.) Now, assume that $m \neq -2$, $m \neq 1$. Set $S = a + b + c$. Adding the given equations gives \[ x + y + z = \frac{S}{m + 2}. \] Hence \[ a = x + y + mz = \frac{S}{m + 2} + (m - 1)z \Leftrightarrow z = \frac{1}{m - 1} \cdot \left(a - \frac{S}{m + 2}\right). \] Similarly \[ y = \frac{1}{m - 1} \cdot \left(b - \frac{S}{m + 2}\right),\ x = \frac{1}{m - 1} \cdot \left(c - \frac{S}{m + 2}\right). \] Now, it becomes obvious that $x,y,z$ are in arithmetic progression if and only if $a,b,c$ are in arithmetic progression. So, that's it.