382. Solve the system of equations:

$$
\left\{\begin{array}{l}
\frac{x y z}{x+y}=\frac{6}{5} \\
\frac{x y z}{y+z}=2 \\
\frac{x y z}{z+x}=\frac{3}{2}
\end{array}\right.
$$
$\triangle$ In each of the equations of the system, we will transition to the inverse quantities and in the left parts of the resulting equations, we will divide the numerators of the fractions term by term by the denominators:
$\frac{x+y}{x y z}=\frac{5}{6}, \quad \frac{y+z}{x y z}=\frac{1}{2}, \quad \frac{z+x}{x y z}=\frac{2}{3} ;$

$\frac{1}{y z}+\frac{1}{x z}=\frac{5}{6}, \quad \frac{1}{x z}+\frac{1}{x y}=\frac{1}{2}, \quad \frac{1}{x y}+\frac{1}{y z}=\frac{2}{3}$.

Let's add all the equations of the last system:

$$
2\left(\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}\right)=2, \quad \frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=1
$$

From here, taking into account the equations of the new system, we will have:

$\frac{1}{x y}=\frac{1}{6}, \quad \frac{1}{y z}=\frac{1}{2}, \quad \frac{1}{z x}=\frac{1}{3} ;$

$x y=6, \quad y z=2, \quad z x=3$.

Let's multiply all the equations of the last system:

$$
x^{2} y^{2} z^{2}=36, \quad x y z= \pm 6
$$

Now it is not difficult to find $x, y$ and $z$.

Answer: $(3 ; 2 ; 1),(-3 ;-2 ;-1)$.