Example 2 Positive numbers $x, y, z$ satisfy the system of equations $\left\{\begin{array}{l}x^{2}+x y+\frac{1}{3} y^{2}=25 \text {, } \\ \frac{1}{3} y^{2}+z^{2}=9, \\ z^{2}+x z+x^{2}=16 .\end{array}\right.$
Find the value of $x y+2 y z+3 z x$.
Solve: Transform the original system of equations into
\[
\left\{
\begin{array}{l}
x^{2}-2 x\left(\frac{\sqrt{3}}{3} y\right) \cos 150^{\circ}+\left(\frac{\sqrt{3}}{3} y\right)^{2}=5^{2}, \\
\left(\frac{\sqrt{3}}{3} y\right)^{2}+z^{2}=3^{2}, \\
z^{2}-2 x z \cos 120^{\circ}+x^{2}=4^{2}.
\end{array}
\right.
\]
From this, by reversing the cosine rule, construct \(\triangle O A B\), \(\triangle O B C\), and \(\triangle O A C\), to form \(\triangle A B C\).
Since \(A C^{2}+B C^{2}=A B^{2}\), it follows that \(\angle A C B=90^{\circ}\),
thus \(6=S_{\triangle A B C}=S_{\triangle O A B}+S_{\triangle O B C}+S_{\triangle O A C}=\frac{1}{4 \sqrt{3}}(x y+2 y z+3 z x)\).
Therefore, \(x y+2 y z+3 z x=24 \sqrt{3}\).