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 $M A=z$,
$$\begin{array}{l}
M B=\frac{\sqrt{3}}{3} y, \\
M C=x
\end{array}$$

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
$$\frac{\sqrt{3}}{12}(2 y z+x y+3 x z)=6 \text {, }$$

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