1. Let point $O$ be outside $\triangle A B C$, and
$$
\overrightarrow{O A}-2 \overrightarrow{O B}-3 \overrightarrow{O C}=0 \text {. }
$$

Then $S_{\triangle A B C}: S_{\triangle O B C}=$
$-1.4$.
As shown in Figure 4, let $D$ and $E$ be the midpoints of sides $AB$ and $BC$, respectively, and connect $CD$. Then
$$
\begin{array}{l}
\overrightarrow{OA}+\overrightarrow{OB}=2 \overrightarrow{OD}, \\
\overrightarrow{OB}+\overrightarrow{OC}=2 \overrightarrow{OE} .
\end{array}
$$
(1) - (2) $\times 3$ gives
$$
\begin{array}{l}
0=\overrightarrow{OA}-2 \overrightarrow{OB}-3 \overrightarrow{OC} \\
=2 \overrightarrow{OD}-6 \overrightarrow{OE} . \\
\text { Therefore, } \overrightarrow{OD}=3 \overrightarrow{OE} .
\end{array}
$$

Thus, $\overrightarrow{OD}$ and $\overrightarrow{OE}$ are collinear, and $|\overrightarrow{OD}|=3|\overrightarrow{OE}|$.
Hence, $|\overrightarrow{DE}|=2|\overrightarrow{OE}|$.
$$
\text { Therefore, } \frac{S_{\triangle BCD}}{S_{\triangle OBC}}=\frac{2}{1}=2, \frac{S_{\triangle ABC}}{S_{\triangle OBC}}=\frac{2 S_{\triangle BCD}}{S_{\triangle OBC}}=4 \text {. }
$$