338. Solve the system of equations:

$$
\left\{\begin{array}{l}
x+y+z=6 \\
x y+y z+z x=11 \\
x y z=6
\end{array}\right.
$$
$\triangle$ Express $x+y$ from the first equation of the system, and $-xy$ from the third:

$$
x+y=6-z, \quad xy=\frac{6}{z}
$$

Now transform the second equation, expressing its left side through $z$:

$$
\begin{aligned}
& xy+z(x+y)=11, \quad \frac{6}{z}+z(6-z)=11 \\
& 6+6z^2-z^3=11z, \quad z^3-6z^2+11z-6=0
\end{aligned}
$$

(Students familiar with Vieta's theorem for a cubic equation can immediately form the last equation.)

By checking the divisors of the constant term of the polynomial on the left side of the cubic equation, we find its roots: $z_1=1, z_2=2, z_3=3$. It is not difficult to find the corresponding values of $x$ and $y$ in each of these three cases.

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