Solve the following system of equations:

$$
\begin{gathered}
\left(x^{2}+11\right) \sqrt{21+y^{2}}=180, \quad\left(y^{2}+21\right) \sqrt{z^{2}-33}=100 \\
\left(z^{2}-33\right) \sqrt{11+x^{2}}=96
\end{gathered}
$$
Introducing the notations

$$
\sqrt{x^{2}+11}=u, \quad \sqrt{y^{2}+21}=v, \quad \sqrt{z^{2}-33}=w
$$

our system of equations takes the following form:

$$
u^{2} v=180, \quad v^{2} w=100, \quad w^{2} u=96
$$

Express $v$ from the first equation, substitute it into the second, then express $w$ from there and substitute it into the third equation:

$$
\frac{1}{v}=\frac{u^{2}}{180}, w=\frac{100}{v^{2}}=\frac{100 u^{4}}{180^{2}}=\frac{u^{4}}{18^{2}}, \frac{u^{9}}{18^{4}}=96
$$

From this,

$$
u=\sqrt[9]{96 \cdot 18^{4}}=\sqrt[9]{2^{9} \cdot 3^{9}}=2 \cdot 3=6
$$

and consequently $v=5, w=4$. Finally, for the original unknowns:

$$
x= \pm \sqrt{u^{2}-11}= \pm 5, \quad y= \pm \sqrt{v^{2}-21}= \pm 2, \quad z= \pm \sqrt{w^{2}+33}= \pm 7
$$

The signs of $x, y, z$ are independent of each other, so by varying the signs, we get 8 solutions:

$$
\begin{aligned}
& x, y, z=+5,+2,+7 ; \quad+5,+2,-7 ; \quad+5,-2,+7 ; \quad+5,-2,-7 \\
& \begin{array}{llll}
-5,+2,+7 ; & -5,+2,-7 ; & -5,-2,+7 ; & -5,-2,-7
\end{array}
\end{aligned}
$$

Zoltán Csörnyei (Veszprém, L. Lovassy II. o. t.)

Remark. Many contestants in the 1st grade did not take into account the two-valued nature of the square root and only provided the first solution above.