Example 1. Given $x, y, z \in R^{+}$, and satisfying
$$
\begin{array}{l}
x^{2}+x y+\frac{y^{2}}{3}=25, \frac{y^{2}}{3}+z^{2}=9, z^{2}+z x+x^{2} \\
=1 \text {, find } x y+2 y z+3 z x \text { . }
\end{array}
$$
Solve: The original system of equations can be transformed into $x^{2}+\left(\sqrt{\frac{y}{3}}\right)^{2}$ $-2 x\left(-\frac{y}{3}\right) \cos 150^{\circ}=5^{2},\left(-\frac{y}{\sqrt{3}}\right)^{2}+z^{2}$ $=3^{2}, z^{2}+x^{2}-2 z x \cos 120^{\circ}=4^{2}$ (inspired by the cosine rule). From this, we can construct $\triangle A B C$ as shown in Figure 1.
Since $A C^{2}=A B^{2}+B C^{2}, \angle A B C=90^{\circ}$,
$$
\begin{array}{c}
\sigma=S_{\triangle \mathrm{ABC}}=\frac{1}{4 \sqrt{3}} .(x y+2 y z+3 z x), \\
\therefore \quad x y+2 y z+3 z x=24 \sqrt{3} .
\end{array}
$$