Example 1 Given $a, b, c > 0$, and
$$
\left\{\begin{array}{l}
a^{2}+a b+\frac{b^{2}}{3}=25, \\
\frac{b^{2}}{3}+c^{2}=9, \\
a^{2}+a b a^{2}+a c+c^{2}=16 .
\end{array}\right.
$$

Try to find the value of $a b+2 b c+3 a c$.
Analysis: It is not easy to solve this problem directly. Observing the numbers on the right side of the equation system is a set of Pythagorean numbers, they can be represented as the three sides of a right triangle, with two sides perpendicular to each other. Therefore, a Cartesian coordinate system can be established, and the solution can be found using the perpendicular relationship of lines and the distance from a point to a line.
Solution: Establish a rectangular coordinate system as shown in Figure 1. Then
$$
\begin{array}{l}
A\left(-\frac{\sqrt{3}}{3} b, c\right), \\
B\left(\frac{\sqrt{3}}{2} a, \frac{a+2 c}{2}\right) .
\end{array}
$$

Given $O A=3$,
$$
\begin{array}{l}
O B=\sqrt{\frac{3}{4} a^{2}+\left(\frac{a^{2}+4 a c+4 c^{2}}{4}\right)} \\
=\sqrt{a^{2}+a c+c^{2}}=4, \\
A B=5, O A^{2}+O B^{2}=A B^{2} .
\end{array}
$$

Therefore, $\triangle A B C$ is a right triangle.
The equation of the line $O A$ is $\frac{\sqrt{3}}{3} b y+c x=0$, and since $O A \perp O B$, the distance from point $B$ to $O A$ is
$$
\begin{array}{l}
\frac{\left|\frac{\sqrt{3}}{3} b \cdot \frac{a+2 c}{2}+c \cdot \frac{\sqrt{3}}{2} a\right|}{\sqrt{\left(\frac{\sqrt{3}}{3} b\right)^{2}+c^{2}}} \\
=4=\frac{|a b+2 b c+3 a c|}{2 \sqrt{3} \times 3} .
\end{array}
$$

Therefore, $a b+2 b c+3 a c=24 \sqrt{3}$.