1. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& x y=a x+a y, \\
& x z=2 x+2 z, \\
& y z=3 y+3 z,
\end{aligned}
$$

where $a$ is a real parameter. Discuss the number of solutions in terms of the parameter $a$.
1. From the second and third equations, which we rewrite in the form

$$
x(z-2)=2 z, \quad y(z-3)=3 z
$$

it follows that $z \notin\{2,3\}$ and the expressions

$$
x=\frac{2 z}{z-2}, \quad y=\frac{3 z}{z-3}
$$

Substituting these into the first equation of the system, we get

$$
\frac{6 z^{2}}{(z-2)(z-3)}=\frac{2 a z}{z-2}+\frac{3 a z}{z-3}
$$

and after simplification,

$$
z((6-5 a) z+12 a)=0
$$

From this, it follows that either $z=0$ (then $x=y=0$), or (assuming $a \neq \frac{6}{5}$)

$$
z=\frac{12 a}{5 a-6}
$$

from which, according to (1), we obtain the solution

$$
x=\frac{12 a}{a+6}, \quad y=\frac{12 a}{6-a}, \quad z=\frac{12 a}{5 a-6}.
$$

The condition $z \notin\{2,3\}$ is equivalent to the condition $a \notin\{-6,6\}$. Additionally, note that for $a=0$, (2) gives the solution $x=y=z=0$. This solution clearly satisfies the system for any real $a$.

Conclusion. For $a \in\left\{-6,0, \frac{6}{5}, 6\right\}$, the system has a unique solution $x=y=z=0$; for other real values of $a$, the system also has a non-zero solution (2).

Alternative Solution. The triplet $x=y=z=0$ is a solution to the given system. From the second and third equations, it follows that if one of the numbers $x, y, z$ is zero, the other two are also zero. Therefore, we further assume that $x y z \neq 0$. Then necessarily $a \neq 0$. We rewrite the system in the form

$$
\frac{1}{a}=\frac{1}{x}+\frac{1}{y}, \quad \frac{1}{2}=\frac{1}{x}+\frac{1}{z}, \quad \frac{1}{3}=\frac{1}{y}+\frac{1}{z}
$$

This is a linear system in terms of the unknowns $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$. We easily find its (unique) solution

$$
\frac{1}{x}=\frac{1}{12}+\frac{1}{2 a}, \quad \frac{1}{y}=-\frac{1}{12}+\frac{1}{2 a}, \quad \frac{1}{z}=\frac{5}{12}-\frac{1}{2 a}.
$$

From this, we determine the triplet

$$
x=\frac{12 a}{a+6}, \quad y=\frac{12 a}{6-a}, \quad z=\frac{12 a}{5 a-6}
$$

which is a solution if $a \notin\left\{-6,6, \frac{6}{5}\right\}$.

Conclusion. For $a \in\left\{-6,0, \frac{6}{5}, 6\right\}$, the system has a unique solution $x=y=z=0$; for other values of $a$, the system also has a second solution (3).

For a complete solution to the problem, award 6 points.