4. 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) $\frac{5}{3}$
(C)
4. 【Analysis and Solution】Solution 1 As shown in the figure, let $D, E$ be the midpoints of sides $A C, B C$ respectively,
then $\overrightarrow{O A}+\overrightarrow{O C}=2 \overrightarrow{O D}$
$$
2(\overrightarrow{O B}+\overrightarrow{O C})=4 \overrightarrow{O E}
$$

From (1) and (2), we get,
$$
\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=2(\overrightarrow{O D}+2 \overrightarrow{O E})=0
$$

Thus, $\overrightarrow{O D}$ and $\overrightarrow{O E}$ are collinear,
and $|\overrightarrow{O D}|=2|\overrightarrow{O E}| \quad \therefore \frac{S_{\triangle A E C}}{S_{\triangle A O C}}=\frac{3}{2}, \therefore \frac{S_{\triangle A B C}}{S_{\triangle A O C}}=\frac{3 \times 2}{2}=3$, hence the answer is C.
Solution 2
$$
\begin{aligned}
& \overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=0 \\
\Rightarrow & \overrightarrow{O A}+2(\overrightarrow{O A}+\overrightarrow{A B})+3(\overrightarrow{O A}+\overrightarrow{A C})=2 \overrightarrow{A B}+6 \overrightarrow{O A}+3 \overrightarrow{A C}=\overrightarrow{0} \\
\Rightarrow & (2 \overrightarrow{A B}+6 \overrightarrow{O A}+3 \overrightarrow{A C}) \times \overrightarrow{A C}=\overrightarrow{0} \Rightarrow|\overrightarrow{A B} \times \overrightarrow{A C}|=3|\overrightarrow{O A} \times \overrightarrow{A C}| \\
\Rightarrow & S_{\triangle A B C}=S_{\triangle A B C}
\end{aligned}
$$

Hence the answer is C.