Example 2. Let positive numbers $a, b, c$ satisfy $\left\{\begin{array}{l}a^{2}+a b+\frac{b^{2}}{3}=25, \\ \frac{b^{2}}{3}+c^{2}=9, \\ c^{2}+a c+a^{2}=16 .\end{array}\right.$ Find the value of $a b+2 b c+3 c c$.
Analysis: The conditional equation resembles the cosine theorem, while the value equation is similar to the sum of areas. Transform it in this direction: 
\[ a^{2}+\left(\frac{b}{\sqrt{3}}\right)^{2}- 2 a\left(\frac{b}{\sqrt{3}}\right) \cos 150^{\circ}=5^{2}, \left(\frac{b}{\sqrt{3}}\right)^{2}+c^{2}=3^{2}, c^{2}+a^{2}- 2 a c \cos 120^{\circ}=4^{2}, \]
This can form a triangle (as shown in the figure). Since
\[ \text { i.e., } \begin{array}{l}
S_{\triangle A B C}=S_{\triangle A O C}+S_{\triangle B O A}+S_{\triangle O O B} \\
\frac{1}{2} \times 3 \times 4=\frac{1}{2} \cdot \cdot \frac{i}{1}=\sin 150^{\circ} \\
\quad \dot{r} \frac{1}{2} \cdot \frac{b}{\sqrt{3}} \cdot c+\frac{1}{2} a c \sin 120^{\circ}, \\
\therefore a b+2 b c+3 a c=24 \sqrt{3} .
\end{array} \]