4. 136 Positive numbers \( x, y, \approx \) satisfy the system of equations
\[
\left\{\begin{array}{l}
x^{2}+x y+\frac{y^{2}}{3}=25, \\
\frac{y^{2}}{3}+z^{2}=9, \\
z^{2}+z x+x^{2}=16 .
\end{array}\right.
\]

Find the value of \( x y + 2 y z + 3 z x \).
[Solution] Take point $M$, and draw line segments $M A, M B$, $M C$, such that $\angle A M B=90^{\circ}, \angle B M C=150^{\circ}$, $\angle C M A=120^{\circ}$. If $M A=z$,
$$
\begin{array}{l}
\text { MB }=\frac{\sqrt{3}}{3} y, \\
M C=x,
\end{array}
$$

then by the cosine rule we get
$$
\begin{aligned}
A B & =\sqrt{M A^{2}+M B^{2}} \\
= & \sqrt{z^{2}+\frac{y^{2}}{3}}=3, \\
B C & =\sqrt{M B^{2}+M C^{2}-2 M B \cdot M C \cos \angle B M C} \\
& =\sqrt{x^{2}+x y+\frac{y^{2}}{3}}=5, \\
A C & =\sqrt{M A^{2}+M C^{2}-2 M A \cdot M C \cos \angle C M A} \\
& =\sqrt{z^{2}+z x+x^{2}}=4 .
\end{aligned}
$$

Thus, $\triangle A B C$ is a right triangle, and
$$
S_{\triangle A B C}=\frac{1}{2} \cdot 3 \cdot 4=6 .
$$

On the other hand,
$$
\begin{aligned}
S_{\triangle A B C} & =S_{\triangle A M B}+S_{\triangle B M C}+S_{\triangle C M A} \\
& =\frac{1}{2} \cdot z \cdot \frac{\sqrt{3}}{3} y+\frac{1}{2} \cdot x \cdot \frac{\sqrt{3}}{3} y \cdot \sin 150^{\circ}+\frac{1}{2} \cdot x z \cdot \sin 120^{\circ} \\
& =\frac{\sqrt{3}}{12}(2 y z+x y+3 x z)
\end{aligned}
$$

From (1) and (2) we get
$$
\frac{\sqrt{3}}{12}(2 y z+x y+3 x z)=6 \text {, }
$$

thus $\quad x y+2 y z+3 z x=24 \sqrt{3}$.