$4 \cdot 127$ Solve the system of equations
$$
\left\{\begin{array}{l}
y z=3 y+2 z-8 \\
z x=4 z+3 x-8 \\
x y=2 x+y-1
\end{array}\right.
$$
[Solution] Rewrite the three equations of the system as
$$
\left\{\begin{array}{l}
(y-2)(z-3)=-2, \\
(z-3)(x-4)=4, \\
(x-1)(y-2)=1 .
\end{array}\right.
$$

From (1) and (3), we get $(z-3)=-2(x-1)$, substituting into (2), we get $(x-3)(x-2)=0$.
That is, $x=2$ or $x=3$.
Substituting back into (2) and (3), we get
$$
\left\{\begin{array}{l}
x=2, \\
y=3, \\
z=1;
\end{array} \quad \left\{\begin{array}{l}
x=3, \\
y=\frac{5}{2}, \\
z=-1 .
\end{array}\right.\right.
$$