1. By subtracting the second equation from the first and the third from the second, we obtain two equations

$$
\begin{aligned}
& (x-y)(x+y-6)=0, \\
& (y-z)(y+z-6)=0,
\end{aligned}
$$

which, together with any of the three given equations, form a system equivalent to the given system. To satisfy the obtained two equations, we have four possibilities:

If $x=y=z$, substituting into any of the given equations yields $y^{2}+12 y-85=0$ and thus $y=5$ or $y=-17$.

If $x=y, z=6-y$, substituting into the first given equation yields $y^{2}+36=85$, and thus $y=7$ or $y=-7$.

If $x=6-y, z=y$, substituting into the last given equation again yields $y^{2}+36=85$, and thus $y=7$ or $y=-7$.

If $x=z=6-y$, substituting into the second given equation yields $y^{2}+6(12-2 y)=85$ or $y^{2}-12 y-13=0$ and thus $y=-1$ or $y=13$.

Answer. The system of equations has eight solutions, namely $(5,5,5),(-17,-17,-17),(7,7,-1)$, $(-7,-7,13),(-1,7,7),(13,-7,-7),(7,-1,7),(-7,13,-7)$.

For a complete solution, award 6 points. For guessing the solution $x=y=z \in\{5,-17\}$, award 2 points, or just 1 point if the solver finds only one of these solutions. Award 2 points for analyzing the other two possibilities (it is the same quadratic equation) and the last possibility $x=z=6-y$ also 2 points. If the student analyzes all four cases but forgets the solutions that arise from changing the order, award only 5 points.
1. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& x^{2}+6(y+z)=85 \\
& y^{2}+6(z+x)=85 \\
& z^{2}+6(x+y)=85
\end{aligned}
$$