～Let point $O$ be inside $\triangle A B C$, and $\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=\overrightarrow{0}$. 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) $\frac{5}{3}$
Given the figure, let $D$ and $E$ be the midpoints of sides $AC$ and $BC$, respectively. Then,
$$
\overrightarrow{OA} + \overrightarrow{OC} = 2 \overrightarrow{OD},
$$
$$
2(\overrightarrow{OB} + \overrightarrow{OC}) = 4 \overrightarrow{OE}
$$

From (1) and (2), we get $\overrightarrow{OA} + 2 \overrightarrow{OB} + 3 \overrightarrow{OC} = 2(\overrightarrow{OD} + 2 \overrightarrow{OE}) = \overrightarrow{0}$, which means $\overrightarrow{OD}$ and $\overrightarrow{OE}$ are collinear, and $|\overrightarrow{OD}| = 2|\overrightarrow{OE}|$.
Therefore,
$$
\frac{S_{\triangle AEC}}{S_{\triangle AOC}} = \frac{3}{2}, \quad \frac{S_{\triangle ABC}}{S_{\triangle AOC}} = \frac{3 \times 2}{2} = 3.
$$

Thus, the correct answer is C.