Example 1 (2004 National High School Competition Question) As shown in Figure 8-1, let point $O$ be inside $\triangle A B C$, and $\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=0$. Then the ratio of the area of $\triangle A B C$ to the area of $\triangle A O C$ is ( ).
A. 2
B. $\frac{3}{2}$
C. 3
D. $\frac{5}{3}$
Solution: Choose C. Reason: Let $M, N$ be the midpoints of $AC, BC$ respectively, then $\overrightarrow{OA}+\overrightarrow{OC}=2\overrightarrow{OM}, 2(\overrightarrow{OB}+\overrightarrow{OC})=4\overrightarrow{ON}$.
Thus, $\overrightarrow{OA}+2\overrightarrow{OB}+3\overrightarrow{OC}=2(\overrightarrow{OM}+2\overrightarrow{ON})=0$, which means $\overrightarrow{OM}$ and $\overrightarrow{ON}$ are collinear, and $|\overrightarrow{OM}|=2|\overrightarrow{ON}|$.
Therefore, $\frac{S_{\triangle ANC}}{S_{\triangle NX}}=\frac{3}{2}$, so $\frac{S_{\triangle ABC}}{S_{\triangle NC}}=\frac{3 \cdot 2}{2}=3$.
Alternative solution: As shown in Figure 8-1, extend $OB$ to $E$ such that $OE=2OB$, and extend $OC$ to $F$ such that $OF=3OC$, then $\overrightarrow{OA}+\overrightarrow{OE}+\overrightarrow{OF}=0$.
Thus, point $O$ is the centroid of $\triangle AEF$. Clearly,
$$
\begin{array}{l}
S_{\triangle NXC}=\frac{1}{3} S_{\triangle MOF}=\frac{1}{9} S_{\triangle AEF}, \\
S_{\triangle NOB}=\frac{1}{2} S_{\triangle NOE}=\frac{1}{6} S_{\triangle AEF}, \\
S_{\triangle BXC}=\frac{1}{6} S_{\triangle AWF}=\frac{1}{18} S_{\triangle AEF} .
\end{array}
$$

Therefore, $S_{\triangle ABC}=\frac{1}{3} S_{\triangle AEF}$, hence $S_{\triangle ABC}=3 S_{\triangle AXC}$.