2. Solve the system of equations

$$
\left\{\begin{array}{l}
11 x-y z=18 \\
11 y-z x=18 \\
11 z-x y=18
\end{array}\right.
$$
Solution. If we subtract the second equation from the first, we sequentially obtain

$$
\begin{aligned}
& 11 x-11 y-y z+z x=0 \\
& 11(x-y)+z(x-y)=0 \\
& (x-y)(11+z)=0
\end{aligned}
$$

There are two possible cases.

1) Let $x=y$. Then, if we subtract the third equation $11 z-x^{2}=18$ from $11 x-x z=18$, we get $(x-z)(11+x)=0$. If $x=z$, then $x=y=z$, so we get $x^{2}-11 x+18=0$, from which we obtain $x=9$ or $x=2$. Thus, in this case, the solutions to the system are $(x, y, z)=(9,9,9)$ and $(x, y, z)=(2,2,2)$. If $11+x=0$, then $x=-11$, so $y=-11$ and from there $z=\frac{x^{2}+18}{11}=\frac{139}{11}$, so the solution to the system is $(x, y, z)=\left(-11,-11, \frac{139}{11}\right)$.
2) Let $z+11=0$, i.e., $z=-11$. Then from the third equation of the system, it follows that $x y=-121-18=-139$, and from the first equation, it follows that $11 x+11 y=18$, i.e., $x+y=\frac{18}{11}$. Thus, $x$ and $y$ are solutions to the equation $u^{2}-\frac{18}{11} u-139=0$, i.e., $x=\frac{139}{11}$, $y=-11$ or $x=-11, y=\frac{139}{11}$.

Finally, all solutions to the given system are:

$$
(2,2,2),(9,9,9),\left(-11,-11, \frac{139}{11}\right),\left(-11, \frac{139}{11},-11\right) \text { and }\left(\frac{139}{11},-11,-11\right)
$$