12.55 Find the integer solutions of the equation $\frac{x y}{z}+\frac{x z}{y}+\frac{y z}{x}=3$.
(26th Moscow Mathematical Olympiad, 1963)
[Solution] First, we find the positive integer solutions of the given equation: $\left(x_{0}, y_{0}, z_{0}\right)$.
Since if $\left(x_{0}, y_{0}, z_{0}\right)$ is a set of natural number solutions to the equation, then $\left(x_{0},-y_{0},-z_{0}\right),\left(-x_{0}, y_{0},-z_{0}\right),\left(-x_{0},-y_{0}, z_{0}\right)$ are also integer solutions of the equation.

The given equation can be transformed into
$$
x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2}=3 x y z,
$$

then
$$
\begin{aligned}
6 x y z & =2 x^{2} y^{2}+2 x^{2} z^{2}+2 y^{2} z^{2} \\
& =\left(x^{2} y^{2}+x^{2} z^{2}\right)+\left(x^{2} y^{2}+y^{2} z^{2}\right)+\left(x^{2} z^{2}+y^{2} z^{2}\right) \\
& \geqslant 2 x^{2} y z+2 x y^{2} z+2 x y z^{2} \\
& =2 x y z(x+y+z) .
\end{aligned}
$$

Thus, we have
$$
x+y+z \leqslant 3 .
$$

If $x, y, z$ are positive integers, then we can only have
$$
x=y=z=1 .
$$

Therefore, the given equation has four sets of solutions:
$$
(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1) .
$$