359*. Solve the system of equations:

$$
\left\{\begin{array}{l}
x^{2}-y z=3 \\
y^{2}-x z=4 \\
z^{2}-x y=5
\end{array}\right.
$$
$\triangle$ Multiply the first equation by $y$, the second by $z$, and the third by $x$ to form a linear equation with $x, y$, and $z$:

$$
x^{2} y-y^{2} z=3 y, \quad y^{2} z-x z^{2}=4 z, \quad x z^{2}-x^{2} y=5 x
$$

After adding these equations, we get:

$$
5 x+3 y+4 z=0
$$

Now multiply the first equation of the original system by $z$, the second by $x$, and the third by $y$, and add the new equations:

$$
x^{2} z-y z^{2}=3 z, \quad x y^{2}-x^{2} z=4 z, \quad y z^{2}-x y^{2}=5 y
$$

From this, we have:

$$
4 x+5 y+3 z=0
$$

From the resulting system of linear equations:

$$
5 x+3 y+4 z=0, \quad 4 x+5 y+3 z=0
$$

express $x$ and $y$ in terms of $z$. For this purpose, each equation can be divided term by term by $z$ (make sure that no solutions of the original system are lost by such division):

$$
5 \frac{x}{z}+3 \frac{y}{z}+4=0, \quad 4 \frac{x}{z}+5 \frac{y}{z}+3=0
$$

Multiply the first equation of the last system by 5, the second by -3, and add these new equations:

$$
13 \frac{x}{z}=-11, \quad \frac{x}{z}=-\frac{11}{13}
$$

It is now easy to find the ratio $\frac{y}{z}: \frac{y}{z}=\frac{1}{13}$. Then

$$
x=-\frac{11}{13} z, \quad y=\frac{1}{13} z
$$

We need to substitute these expressions for $x$ and $y$ into at least the third equation of the original system and find $z$. Then we find $x$ and $y$.

Answer: $(11 / 6 ;-1 / 6 ;-13 / 6),(-11 / 6 ; 1 / 6 ; 13 / 6)$.