14. Let $O$ be a point inside $\triangle A B C$, and
$$
\overrightarrow{A O}=\frac{1}{3} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C} \text {. }
$$

Then $S_{\triangle O A B}: S_{\triangle O B C}=$
14. $\frac{3}{5}$.

From the given, we have $5 \overrightarrow{O A}+4 \overrightarrow{O B}+3 \overrightarrow{O C}=0$.
As shown in Figure 6, extend $O A, O B, O C$ to $O A^{\prime}, O B^{\prime}, O C^{\prime}$, such that
$$
\begin{array}{l}
\left|O A^{\prime}\right|=5|O A|,\left|O B^{\prime}\right|=4|O B|, \\
\left|O C^{\prime}\right|=3|O C| .
\end{array}
$$

Connect $A^{\prime} B^{\prime}, B^{\prime} C^{\prime}, C^{\prime} A^{\prime}$.
Then consider the third and fourth parts:
During the holiday, $n_{1}, n_{2}, \pi$
$n_{3}$ are not adjacent, and 1
Let $m_{1}=n_{1}, m$
Then $\overrightarrow{O A^{\prime}}+\overrightarrow{O B^{\prime}}+\overrightarrow{O C^{\prime}}=\mathbf{0}$.
Therefore, $O$ is the centroid of $\triangle A^{\prime} B^{\prime} C^{\prime}$.
Thus, $S_{\triangle O A^{\prime} B^{\prime}}=S_{\triangle O B^{\prime} C^{\prime}}=S_{\triangle O C^{\prime} 1^{\prime}}$.
$$
\begin{array}{l}
\text { Also, } \frac{S_{\triangle O A B}}{S_{\triangle O A^{\prime} B^{\prime}}}=\frac{O A}{O A^{\prime}} \cdot \frac{O B}{O B^{\prime}}=\frac{1}{20}, \\
\frac{S_{\triangle O B B}}{S_{\triangle O B^{\prime} C^{\prime}}}=\frac{O B}{O B^{\prime}} \cdot \frac{O C}{O C^{\prime}}=\frac{1}{12}, \\
\text { Hence, } \frac{S_{\triangle O A B}}{S_{\triangle O B C}}=\frac{3}{5} .
\end{array}
$$