Subtracting first from second gives $$(x-z)(x+z+2)=0$$Substracting second from third gives $$(y-x)(y+x-1)=0$$. We consider four cases A) $x=y=z$ All equations become $$2x^2=3x\iff x\in\left\lbrace 0,\frac32\right\rbrace$$B) $x=z,\ x+y=1$ All equations become $$2x^2+x-2=0\iff x\in\left\lbrace \frac{-1-\sqrt{17}}{4},\frac{-1+\sqrt{17}}{4}\right\rbrace$$C) $x+z=-2,\ x=y$ All equations become $$2x^2-x+2=0\iff x\in \emptyset$$D) $x+z=-2,\ x+y=1$ All equations become$$2x^2+3x=0\iff x\in\left\lbrace 0,-\frac32\right\rbrace$$

So the given equations are: $x^2+y^2=-x+3y+z...(1)$ $y^2+z^2=x+3y-z...(2)$ $x^2+z^2=2x+2y-z...(3)$ Given that $x,y,z\in\mathbb{R}$, such that $x\neq y, y\neq z$ and $z\neq x$. Now $(1)-(2)$ gives us: $(x+z)(x-z)=-2(x-z)$ $\implies x+z=-2 (\because x-z\neq 0)$. Now $(2)-(3)$ gives us: $(y+x)(y-x)=y-x$ $\implies y+x=1 (\because y-x\neq 0)$. Therefore $z=-2-x$ and $y=1-x$. Substituting the values of $y$ and $z$ in $(1)$, we have: $$2x^2+3x=x(2x+3)=0$$$$\implies x=0, \frac{-3}{2}.$$ Therefore, $$(x,y,z)=(0,1,-2),\left(\frac{-3}{2},\frac{5}{2},\frac{-1}{2}\right)$$are the required solutions to the system of equations.