Example 12 Given positive numbers $x, y, z$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{2}+x y+\frac{y^{2}}{3}=25, \\
\frac{y^{2}}{3}+z^{2}=9, \\
z^{2}+x z+x^{2}=16 .
\end{array}\right.
$$

Find the value of $x y+2 y z+3 x z$.
(18th All-Union Mathematical Olympiad for Secondary Schools)
The original system of equations is equivalent to
$$
\left\{\begin{array}{l}
x^{2}-2 x \cdot \frac{y}{\sqrt{3}} \cos 150^{\circ}+\left(\frac{y}{\sqrt{3}}\right)^{2}=5^{2}, \\
\left(\frac{y}{\sqrt{3}}\right)^{2}+z^{2}=3^{2}, \\
z^{2}-2 x \cos 120^{\circ}+x^{2}=4^{2} .
\end{array}\right.
$$

Notice that $3^{2}+4^{2}=5^{2}$. As shown in Figure 1, construct a right triangle $\triangle ABC$ and find a point $O$ inside it such that $\angle B O C=90^{\circ}, \angle A O C=120^{\circ}, \angle A O B=150^{\circ}$. Let $O A=x, O B=\frac{y}{\sqrt{3}}$, and $O C=z$. By the Pythagorean theorem and the cosine rule, we know that $O A, O B$, and $O C$ satisfy the system of equations.
$$
\begin{array}{l} 
\because S_{\triangle A B C}=S_{\triangle A K C^{\circ}}+S_{\triangle B B X^{\circ}}+S_{\triangle A O B 3}, \\
\therefore \frac{1}{2} \times 3 \times 4= \frac{1}{2} x z \sin 120^{\circ}+\frac{1}{2} \cdot \frac{y}{\sqrt{3}} \cdot z \\
+\frac{1}{2} x \cdot \frac{y}{\sqrt{3}} \cdot \sin 150^{\circ} .
\end{array}
$$

Simplifying and rearranging, we get
$$
x y+2 y z+3 x z=24 \sqrt{3} \text {. }
$$