5. It is known that positive numbers $x, y, z$ satisfy the system:

$$
\left\{\begin{array}{c}
x^{2}+y^{2}+x y=529 \\
x^{2}+z^{2}+\sqrt{3} x z=441 \\
z^{2}+y^{2}=144
\end{array}\right.
$$

Find the value of the expression $\sqrt{3} x y+2 y z+x z$.
# Solution:

Consider triangle $ABC$ with a point $O$ chosen inside it such that $AO=y, OB=z, OC=x$, $\angle AOB=90^{\circ}, \angle AOC=120^{\circ}, \angle BOC=150^{\circ}$ (see figure). The conditions of the system represent the cosine theorem (including the Pythagorean theorem) for triangles $AOB, AOC, BOC$.

From this, it is not difficult to understand that $AB=12, BC=21$, $AC=23$. Now notice that

![](https://cdn.mathpix.com/cropped/2024_05_06_2ac61c6c474dda84ee57g-2.jpg?height=445&width=680&top_left_y=1362&top_left_x=1267)

$$
\sqrt{3} xy + 2yz + xz = 4S_{AOC} + 4S_{AOB} + 4S_{BOC} = 4(S_{AOC} + S_{AOB} + S_{BOC}) = 4S_{ABC}
$$

The area of triangle $ABC$ will be found using Heron's formula:

$$
p = \frac{12 + 21 + 23}{2} = 28, \quad S_{ABC} = 56\sqrt{5}.
$$

Therefore, $4S_{ABC} = 224\sqrt{5}$.

Answer: $224\sqrt{5}$.