4. Let point $O$ be inside $\triangle A B C$, and $\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\mathbf{0}$. 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. $\frac{5}{3}$
As shown in the figure, let the midpoints of sides $B C$ and $A C$ be $D$ and $E$, respectively. Then we have $\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\overrightarrow{O A}+\overrightarrow{O C}+2(\overrightarrow{O B}+\overrightarrow{O C})$ $=2 \overrightarrow{O E}+4 \overrightarrow{O D}=\mathbf{0} \Rightarrow O$ is the trisection point of segment $D E$. Therefore, $S_{\triangle A O C}=\frac{2}{3} S_{\triangle A D C}=\frac{1}{3} S_{\triangle A B C}$. Hence, the correct choice is $C$.