7. (2nd "Friendship Cup" International Mathematics Competition Problem) Solve the system of equations
$$
\left\{\begin{array}{l}
y z=3 y+2 z-8, \\
z x=4 z+3 x-8, \\
x y=2 x+y-1 .
\end{array}\right.
$$
7. Rewrite the three equations of the system as
$$
\left\{\begin{array}{l}
(y-2)(z-3)=-2, \\
(z-3)(x-4)=4, \\
(x-1)(y-2)=1 .
\end{array}\right.
$$
(2) From (1) and (3), we get $z-3=-2(x-1)$.

Substituting into (2) yields $(x-3)(x-2)=0$, i.e., $x=2$ or $x=3$. Substituting back into (2) and (3) gives the solutions $(x, y, z)$ as $(2,3,1),\left(3, \frac{5}{2},-1\right)$.