In the diagram, let $D$ and $E$ be the midpoints of the sides $AC$ and $BC$, respectively. Then we have
$$\overrightarrow{OA} + \overrightarrow{OC} = 2 \overrightarrow{OD}, \qquad (1)$$
and
$$2(\overrightarrow{OB} + \overrightarrow{OC}) = 4 \overrightarrow{OE}. \qquad (2)$$
By equations (1) and (2) we get
$$\overrightarrow{OA} + 2\overrightarrow{OB} + 3\overrightarrow{OC} = 2(\overrightarrow{OD} + 2\overrightarrow{OE}) = 0.$$
It follows that $\overrightarrow{OD}$ and $\overrightarrow{OE}$ are collinear, and
$|\overrightarrow{OD}| = 2 |\overrightarrow{OE}|$. Consequently, $\frac{S_{\triangle AEC}}{S_{\triangle AOC}} = \frac{3}{2}$
and $\frac{S_{\triangle ABC}}{S_{\triangle AOC}} = \frac{3 \times 2}{2} = 3$. Answer: C.