3. (3 points) Solve the system of equations: $\left\{\begin{array}{l}x y=6(x+y) \\ x z=4(x+z) \\ y z=2(y+z)\end{array}\right.$
Answer: $x=y=z=0$ or $x=-24, y=\frac{24}{5}, y=\frac{24}{7}$

Solution: It is easy to verify that if one of the variables is 0, then the others are also 0. Otherwise, the system can be transformed into

$\left\{\begin{array}{l}\frac{1}{6}=\frac{1}{x}+\frac{1}{y} \\ \frac{1}{4}=\frac{1}{x}+\frac{1}{z} \\ \frac{1}{2}=\frac{1}{z}+\frac{1}{y}\end{array}\right.$

By adding the first two equations and subtracting the last one, we get $\frac{1}{6}+\frac{1}{4}-\frac{1}{2}=\frac{2}{x}$, from which $-\frac{1}{12}=\frac{2}{x}$, that is, $x=-24$. Similarly, we find the other unknowns.