Example 29 (2004 National Competition Question) As shown in Figure 1-11, 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: Given that $O$ is inside $\triangle ABC$ and $\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=0$, this is a very elegant condition, indicating that point $O$ should be the centroid of a triangle (or from the perspective of force balance). Extend $OB$ to $B_{1}$ such that $OB_{1}=2OB$, and extend $OC$ to $C_{1}$ such that $OC_{1}=3OC$. Connect $AB_{1}$, $B_{1}C_{1}$, and $AC_{1}$. Clearly, $O$ is the centroid of $\triangle AB_{1}C_{1}$ (since $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=0$).

At this point, $S_{\triangle ANB_{1}}=S_{\triangle B_{1}(x_{1}}=S_{\triangle AC_{1}}=\frac{1}{3} S_{\triangle AB_{1}c_{1}}$.
And $S_{\triangle AXC}=\frac{1}{3} S_{\triangle AC_{1}}=\frac{1}{9} S_{\triangle AB_{1}c_{1}}$,
$S_{\triangle AOB}=\frac{1}{2} S_{\triangle AOB_{1}}=\frac{1}{6} S_{\triangle AB_{1}C_{1}}$,
$S_{\triangle CBC}=\frac{1}{2} \cdot \frac{1}{3} S_{\triangle B_{1}(C_{1}}=\frac{1}{18} S_{\triangle AB_{1}C_{1}}$, thus
$S_{\triangle ABC}=S_{\triangle AXC}+S_{\triangle AMB}+S_{\triangle YBC}=\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{18}\right) S_{\triangle AB_{1}C_{1}}=\frac{1}{3} S_{\triangle AB_{1}C_{1}}$,

Therefore, $\frac{S_{\triangle ABC}}{S_{\triangle AMC}}=\frac{\frac{1}{3} S_{\angle, AB_{1}C_{1}}}{\frac{1}{9} S_{AB_{1}C_{1}}}=3$.