334. Solve the system of equations:

$$
\left\{\begin{array}{l}
x^{2}+y z=y+z \\
y^{2}+x z=x+z \\
z^{2}+x y=x+y
\end{array}\right.
$$
$\triangle$ Subtract the second equation from the first equation of the system, and the third equation from the second:

$$
\begin{array}{ll}
x^{2}-y^{2}-z(x-y)=y-x, & y^{2}-z^{2}-x(y-z)=z-y \\
(x-y)(x+y-z+1)=0, & (y-z)(y+z-x+1)=0
\end{array}
$$

Now we need to consider four cases, depending on which of the factors in the left-hand side of the last two equations is zero. Work through these cases on your own.

Answer: $(0 ; 0 ; 0),(1 ; 1 ; 1),(-1 ; 1 ; 1),(1 ;-1 ; 1),(1 ; 1 ;-1)$.