Solve as shown in Figure 4, construct
a right-angled triangle with side lengths of 3,
4, and 5, and
make $\angle B P C=90^{\circ}, \angle A P C$
$=150^{\circ}$, then $\angle A P B=$
$120^{\circ}$. By the cosine rule, we get
$$\begin{array}{l} \left\{\begin{array}{l} x^{2}+\left(\frac{y}{\sqrt{3}}\right)^{2}-2 x \cdot \frac{y}{\sqrt{3}} \cos 150^{\circ}=5^{2}, \\ \left(\frac{y}{\sqrt{3}}\right)^{2}+z^{2}-2 \cdot \frac{y}{\sqrt{3}} \cdot z \cdot \cos 90^{\circ}=3^{2}, \\ z^{2}+x^{2}-2 x z \cdot \cos 120^{\circ}=4^{2} . \end{array}\right. \\ \text { Also, } \because S_{\triangle A P B}=\frac{1}{2} x z \sin 120^{\circ}=\frac{\sqrt{3}}{4} x z, \\ S_{\triangle B P C}=\frac{1}{2} \cdot \frac{y}{\sqrt{3}} \cdot z \sin 90^{\circ}=\frac{\sqrt{3}}{6} y z, \\ S_{\triangle A P C}=\frac{1}{2} x \cdot \frac{y}{\sqrt{3}} \sin 150^{\circ}=\frac{\sqrt{3}}{12} x y, \\ S_{\triangle A B C}=\frac{1}{2} \times 4 \times 3=6 . \end{array}$$

And $S_{\triangle A B C}=S_{\triangle A P B}+S_{\triangle A A^{\prime}}+S_{\triangle A P C}$.
$$\begin{array}{l} \therefore \frac{\sqrt{3}}{12} x y+\frac{\sqrt{3}}{6} y z+\frac{\sqrt{3}}{4} x z=6 . \\ \frac{\sqrt{3}}{12}(x y+2 y z+3 x z)=6 . \\ \therefore x y+2 y z+3 x z=24 \sqrt{3} . \end{array}$$