Solve the following system of equations:

$$
\begin{gathered}
x^{2}+y^{2}+z^{2}=14, \\
x y+x z+y z=11, \\
x y z=6 .
\end{gathered}
$$
The second equation, when doubled and added to the first, yields $(x+y+z)^{2}=36$, from which

$$
|x+y+z|=6
$$

It is clear that the system of equations $(1)-(2)-(3)$ is equivalent to the system $\left(1^{\prime}\right)-(2)-(3)$, so it is sufficient to solve the latter. (1) means that either $x+y+z=6$ or $x+y+z=-6$. In the first case, multiplying (2) by $x$ gives

$$
x^{2} y+x^{2} z+x y z=11 x
$$

From (3) and using $y+z=6-x$, we get

$$
\begin{gathered}
x^{2}(6-x)+6=11 x \\
x^{3}-6 x^{2}+11 x-6=(x-1)(x-2)(x-3)=0
\end{gathered}
$$

Thus, the possible values of $x$ are 1, 2, and 3, and by choosing $y$ and $z$ such that $y z=6 / x, y+z=6-x$ holds, (3) and $\left(1^{\prime}\right)$ will automatically be true, and due to $y+z=6-x$, $\left(2^{\prime}\right)$ will also be true, and then (2) will be true because $x \neq 0$. In this case, we obtain the following solution:

| $\mathrm{x}$ | 1 | 1 | 2 | 2 | 3 | 3 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{y}$ | 2 | 3 | 1 | 3 | 1 | 2 |
| $\mathrm{z}$ | 3 | 2 | 3 | 1 | 2 | 1 |

In the case of $x+y+z=-6$, one of $x, y, z$ must be negative, but since $x y z=6$, exactly two must be negative. In this case, two of the addends in (2) will be negative, so the product of the two negative numbers among $x, y$, and $z$ will be at least 11. Consequently, their sum of squares will be greater than 22, so (1) cannot hold. Thus, there is no solution in this case.

In the end, the system of equations (1)-(3) has six solutions, which we have summarized in the above table.