Answer: $(-4 ; 2 ; 1)$.

Solution. Multiplying both sides of the first equation by $12 x(y+z)$, both sides of the second by $6 y(x+z)$, and the third by $2 z(x+y)$, we obtain the system

$$
\left\{\begin{array}{l}
12(x+y+z)=x y+x z \\
6(x+y+z)=x y+y z \\
2(x+y+z)=x z+y z
\end{array}\right.
$$

Adding all three equations term by term and dividing the resulting equality by two, we get the equality

$$
x y+x z+y z=10(x+y+z)
$$

Subtracting each of the equations of the last system from it, we find that

$$
\left\{\begin{array}{l}
8(x+y+z)=x y \\
-2(x+y+z)=y z \\
4(x+y+z)=x z
\end{array}\right.
$$

Dividing the first equation by the second (this is possible because from the domain of the original system it follows that $x y z \neq 0$), we get that $x=-4 z$, and dividing the first by the third - that $y=2 z$.

Then the second equation takes the form $2 z=2 z^{2}$, from which $z=1, x=-4, y=2$.
3. Solve the system of equations $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y+z}=\frac{1}{12} \\ \frac{1}{y}+\frac{1}{x+z}=\frac{1}{6} \\ \frac{1}{z}+\frac{1}{x+y}=\frac{1}{2}\end{array}\right.$,