380. Solve the system of equations:

$$
\left\{\begin{array}{l}
x=\frac{y+1}{3 y-5} \\
y=\frac{3 z-2}{2 z-3} \\
z=\frac{3 x-1}{x-1}
\end{array}\right.
$$
$\triangle$ Substitute the expression for $z$ from the third equation of the system into the second:

$$
\begin{aligned}
& y=\left(3 \frac{3 x-1}{x-1}-2\right) /\left(2 \frac{3 x-1}{x-1}-3\right) \\
& y=\frac{9 x-3-2 x+2}{6 x-2-3 x+3}, \quad y=\frac{7 x-1}{3 x+1}
\end{aligned}
$$

Substitute the last expression for $y$ into the first equation:

$$
\begin{aligned}
& x=\left(\frac{7 x-1}{3 x+1}+1\right) /\left(3 \frac{7 x-1}{3 x+1}-5\right) \\
& x=\frac{7 x-1+3 x+1}{21 x-3-15 x-5}, \quad x=\frac{10 x}{6 x-8} \\
& x=\frac{5 x}{3 x-4}, \quad 3 x^{2}-4 x=5 x, \quad 3 x^{2}-9 x=0
\end{aligned}
$$

From here, $x_{1}=0, x_{2}=3$. Next, we find $y$ and $z$.

Answer: $(0, -1, 1), (3, 2, 4)$.