Example 2 Suppose point $O$ is inside $\triangle A B C$, and $\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=0$.
Then $S_{\triangle A B C}: S_{\triangle A O C}=(\quad$.
(A) 2
(B) $\frac{3}{2}$
(C) 3
(D) $\frac{5}{3}$
(2004, National High School Mathematics Competition)
Let $e$ be the unit normal vector of $A C$, and construct $B D \perp A C$ at point $D$, and $O E \perp A C$ at point $E$, as shown in the figure.
2. Then
$$
\begin{aligned}
D B & =\overrightarrow{A B} \cdot e \\
& =(\overrightarrow{A O}+\overrightarrow{O B}) \cdot e \\
& =\left(\overrightarrow{A O}+\frac{\overrightarrow{A O}+3 \overrightarrow{C O}}{2}\right) \cdot e \\
& =E O+\frac{E O+3 E O}{2}=3 E O .
\end{aligned}
$$

Therefore, $S_{\triangle B C C}: S_{\triangle 1 O C}=3: 1$.
Hence, the correct choice is (C).