Determine all integers $x,\ y,\ z$ that satisfy $x+y+z=(x-y)^{2}+(y-z)^{2}+(z-x)^{2}$

Let $y=u+x$ and $z=v+x$, then $y-z = u-v$ and hence $$ \begin{align} (y-u) + y + (y-u+v) &= u^2 + (u-v)^2 + v^2\\ 3y &= 2u^2 + 2v^2 - 2uv +2u - v \end{align} $$ So we can almost freely choose $u,v$, but we need RHS to be divisible by $3$. By considering the equation modulo $3$, we see that solutions are $$ (u,v) \equiv (0,0),(0,2),(1,1),(1,2),(2,0),(2,1) \pmod 3 $$ For example, taking $(u,v)\equiv (0,0)\pmod 3$, this gives parametrizations: $$ \begin{align} u = 3m, v = 3n &\implies y= 2 m + 6 m^2 - n - 6 m n + 6 n^2\\ &\implies x = -m + 6 m^2 - n - 6 m n + 6 n^2\\ &\implies z = -m + 6 m^2 + 2n - 6 m n + 6 n^2 \end{align} $$ where we can freely choose $m,n\in\mathbb Z$. The other cases are the same.

Edit 1: One more case: let $(u,v)\equiv (2,1)\pmod 3$, so that $$ u = 3m+2,\quad v = 3n+1 $$ then $$ \begin{align} y &= 3 + 8 m + 6 m^2 - n - 6 m n + 6 n^2\\ x = y-u &= 1 + 5 m + 6 m^2 - n - 6 m n + 6 n^2\\ z =v+x &= 2 + 5 m + 6 m^2 + 2 n - 6 m n + 6 n^2 \end{align} $$