Example 4. Positive numbers $x, y, z$ satisfy the system of equations
$$\left\{\begin{array}{l}
x^{2}+x y+\frac{y^{2}}{3}=25 \\
\frac{y^{2}}{3}+z^{2}=9 \\
x^{2}+x y+z^{2}=16
\end{array}\right.$$

Try to find: $x y+2 y z+3 z x$.
Notice that $9=3^{2}, 16=4^{2}, 25=5^{2}$, then we can form a right-angled triangle $\triangle ABC$ with side lengths $3, 4, 5$. Inside this triangle, find a point such that the angles formed by its connections to the three vertices are $90^{\circ}, 120^{\circ}, 150^{\circ}$ (as shown in the figure). Let
$$P B=x, P C=z, P A=\frac{y}{\sqrt{3}} \text {. }$$

Clearly, the sides of $\triangle P A B$,
$\triangle P A C, \triangle P B C$ satisfy the three equations in the given system (using the cosine rule).

By $S_{\triangle A B C}=S_{\triangle P A B}+$ $S_{\triangle P A C}+S_{\triangle P B C}$,
i.e., $\frac{1}{2} \cdot 3 \cdot 4=\frac{1}{2}\left(x \cdot \frac{y}{\sqrt{3}} \sin 150^{\circ}\right.$
$$\left.+\frac{y}{\sqrt{3}} \cdot z+x z \sin 120^{\circ}\right),$$

Thus, $x y+2 y z+3 x z=24 \sqrt{3}$.