Solve the system of equations:

$$
\left\{\begin{array}{l}
\frac{x^{2}+y^{2}-z^{2}}{x y}=\frac{3}{z} \\
\frac{y^{2}+z^{2}-x^{2}}{x y}=\frac{3}{x} \\
\frac{y^{2}-x^{2}-z^{2}}{x z}=\frac{21}{y}
\end{array}\right.
$$
Let's write the first two equations as $z^{2}\left(x^{2}+y^{2}-z^{2}\right)=3 x y z, x^{2}\left(y^{2}+z^{2}-x^{2}\right)=3 x y z$. After subtracting, we get $z^{2} y^{2}-x^{2} y^{2}+x^{4}-z^{4}=0 \Leftrightarrow\left(z^{2}-x^{2}\right)\left(y^{2}-z^{2}-x^{2}\right)=0 \Leftrightarrow y^{2}-z^{2}-x^{2}=0$ or $x^{2}=z^{2}$. The first case contradicts the third equation of the system.

Thus, $z= \pm x$. After substituting into the first two equations, we get $y= \pm 3$, and the third equation will become $y\left(y^{2}\right.$ $\left.-2 x^{2}\right)= \pm 21 x^{2}$. After substituting the value of $y$, we get $3\left(9-2 x^{2}\right)=21 x^{2} \Leftrightarrow x^{2}=1$.

## Answer

$(1,3,1),(-1,3,-1),(1,-3,-1),(-1,-3,1)$