327. Solve the system of equations:

$$
\left\{\begin{array}{l}
y+z=x y z \\
z+x=x y z \\
x+y=x y z
\end{array}\right.
$$
$\triangle$ Subtract the first and second, as well as the first and third equations of the system:

$$
y-x=0, \quad z-x=0
$$

From here, $x=y=z$. Then the first equation reduces to $2 x=x^{3}$. Therefore, $x_{1}=0, x_{2,3}= \pm \sqrt{2}$.

Answer: $(0 ; 0 ; 0),(\sqrt{2} ; \sqrt{2} ; \sqrt{2}),(-\sqrt{2} ;-\sqrt{2} ;-\sqrt{2})$.