Example 12. Positive numbers $x, y, z$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{2}+x y+\frac{1}{3} y^{2}=25, \\
\frac{1}{3} y^{2}+z^{2}=9, \\
z^{2}+x z+x^{2}=16 .
\end{array}\right.
$$

Find the value of $x y+2 y z+3 z x$.
(18th All-Russian High School Mathematics Competition)
Solve: Transform the original system of equations into
$$
\left\{\begin{array}{l}
x^{2}+\left(\frac{y}{\sqrt{3}}\right)^{2}-2 \cdot x \cdot \frac{y}{\sqrt{3}} \cdot \cos 150^{\circ}=5^{2}, \\
\left(\frac{y}{\sqrt{3}}\right)^{2}+z^{2}=3^{2}, \\
z^{2}+x^{2}-2 x z \cos 120^{\circ}=4^{2} .
\end{array}\right.
$$

From this, by reversing the cosine rule, construct $\triangle O A B, \triangle O B C$, and $\triangle O C A$, to get $\triangle A B C$.
$$
\begin{array}{l}
\because A C^{2}+B C^{2} \\
\quad=A B^{2}, \\
\therefore \angle A C B=90^{\circ}, \\
\therefore 6=S_{\triangle A B C} \\
=S_{\triangle O A B}+S_{\triangle O B C} \\
\quad+S_{\triangle O A C} \\
\quad=\frac{1}{4 \sqrt{3}}(x y+2 y z+3 z x) .
\end{array}
$$

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