[Algebra]
Problem: Determine all triples $(x, y, z)$ of real numbers that satisfy the following system of equations:

$$
\begin{aligned}
(x-1)(y-2) & =0 \\
(x-3)(z+2) & =0 \\
x+y z & =9
\end{aligned}
$$

Solution: Since $(x-1)(y-2)=0$, then $x=1$ or $y=2$.

Suppose that $x=1$. In this case, the remaining equations become:

$$
\begin{aligned}
(1-3)(z+2) & =0 \\
1+y z & =9
\end{aligned}
$$

or

$$
\begin{array}{r}
-2(z+2)=0 \\
y z=8
\end{array}
$$

From the first of these equations, $z=-2$.

From the second of these equations, $y(-2)=8$ and so $y=-4$.

Therefore, if $x=1$, the only solution is $(x, y, z)=(1,-4,-2)$.

Suppose that $y=2$. In this case, the remaining equations become:

$$
\begin{aligned}
(x-3)(z+2) & =0 \\
x+2 z & =9
\end{aligned}
$$

From the first equation $x=3$ or $z=-2$.

If $x=3$, then $3+2 z=9$ and so $z=3$.

If $z=-2$, then $x+2(-2)=9$ and so $x=13$.

Therefore, if $y=2$, the solutions are $(x, y, z)=(3,2,3)$ and $(x, y, z)=(13,2,-2)$.

In summary, the solutions to the system of equations are

$$
(x, y, z)=(1,-4,-2),(3,2,3),(13,2,-2)
$$

We can check by substitution that each of these triples does indeed satisfy each of the equations.

Final answer: (1,-4,-2),(3,2,3),(13,2,-2)