3. Solve the system of equations

$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=19^{2} \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \\
x-y+z=11
\end{array}\right.
$$
Solution. The given system is equivalent to the system

$$
\left\{\begin{array}{l}
(x+y+z)^{2}-2(x y+y z+z x)=19^{2} \\
x y+y z+z x=0 \\
x-y+z=11
\end{array}\right.
$$

from which we obtain

$$
\left\{\begin{array}{l}
(x+y+z)^{2}=19^{2} \\
x-y+z=11
\end{array}\right.
$$

The latter system is equivalent to the following two systems

$$
\begin{aligned}
& \left\{\begin{array}{l}
x+z=15 \\
y=4
\end{array}\right. \\
& \left\{\begin{array}{l}
x+z=-4 \\
y=-15
\end{array}\right.
\end{aligned}
$$

From system (1) and any of the equations of the given system, we obtain the following two solutions: $\left(\frac{15+\sqrt{465}}{2}, 4, \frac{15-\sqrt{465}}{2}\right)$ and $\left(\frac{15-\sqrt{465}}{2}, 4, \frac{15+\sqrt{465}}{2}\right)$.

From system (2) and any of the equations of the given system, the following two solutions are obtained: $(6,-15,-10)$ and $(-10,-15,6)$.