Example 15. Given positive numbers $x, y, z$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{2}+x y+\frac{1}{3} y^{2}=25, \\
z^{2}+\frac{1}{3} y^{2}=9, \\
x^{2}+x z+z^{2}=16 .
\end{array}\right.
$$

Find the value of $x y+2 y z+3 x z$.
Solve the original system of equations, i.e.,
$$
\left\{\begin{array}{l}
x^{2}+\left(\frac{y}{\sqrt{3}}\right)^{2}-2 x \\
\cdot \frac{y}{\sqrt{3}} \cos 150^{\circ}=5^{2}, \\
z^{2}+\left(\frac{y}{\sqrt{3}}\right)^{2}=3^{2}, \\
x^{2}+z^{2}-2 x z \cos 120^{\circ}=4^{2} .
\end{array}\right.
$$

From this, we can construct
$\mathrm{Rt} \triangle A B C$, as shown in the figure.
$$
\begin{array}{c}
\because S_{\triangle A P B}+ \\
S_{\triangle B P C}+S_{\triangle C P A}=S_{\triangle A B C}, \\
\text { i.e., } \frac{1}{2} x \cdot \frac{y}{\sqrt{3}} \sin 150^{\circ} \\
+\frac{1}{2} z \cdot \frac{y}{\sqrt{3}} \\
+\frac{1}{2} x z \sin 120^{\circ}=\frac{1}{2} \times 3 \times 4, \\
\quad \therefore x y+2 y z+3 x z=24 \sqrt{3} .
\end{array}
$$