3. Solve the system of equations:

$$
\left\{\begin{array}{l}
x y+5 y z-6 x z=-2 z \\
2 x y+9 y z-9 x z=-12 z \\
y z-2 x z=6 z
\end{array}\right.
$$
Answer: $(-2,2,1 / 6),(x, 0,0),(0, y, 0)$, where $\boldsymbol{x}, \boldsymbol{y}$ are any real numbers.

Solution: From the third equation of the system, we find $y z-2 x z-6 z=0$, from which $z(y-2 x-6)=0$. Therefore, $z=0$ or $y=2 x+6$.

1) Let $z=0$, then the first and second equations reduce to the equality $x y=0$. Therefore, $x=0$ or $y=0$. Thus, the solutions are all possible triples of numbers of the form $(x, 0,0),(0, y, 0)$, where $x, y$ are any real numbers.
2) Let $z \neq 0, y=2 x+6$. Subtracting twice the first equation of the original system from the second, we get $3 x z-y z=-8 z$. Since $z \neq 0$, then $y=3 x+8$. Considering the condition $y=2 x+6$, we have $x=-2, y=2$. From the first or second equation of the original system, we then find $z=1 / 6$.