2. (2004 National High School Mathematics League) Let point $O$ be inside $\triangle A B C$, and $\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\mathbf{0}$. Find the ratio of the area of $\triangle A B C$ to the area of $\triangle A O C$.
2. Extend $O B$ to $O B^{\prime}$ such that $\left|\overrightarrow{O B^{\prime}}\right|=2|\overrightarrow{O B}|$, and extend $C$ to $C^{\prime}$ such that $\left|\overrightarrow{O C^{\prime}}\right|=3|\overrightarrow{O C}|$. By vector knowledge, $O$ is the centroid of $\triangle A B^{\prime} C^{\prime}$, so $S_{A O B^{\prime}}=S_{\triangle B^{\prime}\left(C^{\prime}\right.}=S_{\triangle C O A}$. Let's assume each area is 1, then
$$
\frac{S_{\triangle A O B}}{S_{\triangle A O B^{\prime}}}=\frac{\frac{1}{2} A O \cdot B O \cdot \sin \angle A O B}{\frac{1}{2} A O \cdot B^{\prime} O \cdot \sin \angle A O B}=\frac{1}{2}
$$

So $S_{A(B B}=\frac{1}{2}$ Similarly: $S_{B(X}=\frac{1}{6}, S_{\alpha), A}=\frac{1}{3}$
Therefore, $S_{\triangle B C}=1$ So $\frac{S_{\triangle A B C}}{S_{\triangle A X:}}=3$.