2. Find the non-negative real solutions of the system of equations:
$$
\left\{\begin{array}{l}
x+y+z=3 x y, \\
x^{2}+y^{2}+z^{2}=3 x z, \\
x^{3}+y^{3}+z^{3}=3 y z
\end{array}\right.
$$
Solving, we get $x=z=0$. Therefore, $x=y=z=0$.
If $y \neq 0$, then let $a=\frac{x}{y}, b=\frac{z}{y}$, eliminate $x, y$, and set $m=a+b, n=ab$. Solving, we get $m=2, n=1$. Thus, $a=b=1 \Rightarrow x=y=z$. Therefore, the unique non-negative real solution to the original system of equations is $(x, y, z)=(1,1,1)$. 
Hint: If $y=0$, then $x+z=0, x^{2}+z^{2}=3 x z$.