3. Let point $P$ be inside $\triangle A B C$, and
$$
\begin{array}{l}
\overrightarrow{A P}=\frac{1}{3} \overrightarrow{P B}+\frac{1}{4} \overrightarrow{P C} . \\
\text { Then } \frac{S_{\triangle P B C}}{S_{\triangle A B C}}=
\end{array}
$$
3. $\frac{12}{19}$.

As shown in Figure 2, extend $A P$ and $B C$ to intersect at point $D$.

From $\overrightarrow{A P}$
$$
\begin{array}{l}
= \frac{1}{3}(\overrightarrow{P D}+\overrightarrow{D B})+ \\
\frac{1}{4}(\overrightarrow{P D}+\overrightarrow{D C}) \\
= \frac{7}{12} \overrightarrow{P D}+\left(\frac{1}{3} \overrightarrow{D B}+\frac{1}{4} \overrightarrow{D C}\right) \\
\Rightarrow \frac{1}{3} \overrightarrow{D B}+\frac{1}{4} \overrightarrow{D C}=\mathbf{0}, \overrightarrow{A P}=\frac{7}{12} \overrightarrow{P D} . \\
\text { Therefore, } \frac{S_{\triangle P B C}}{S_{\triangle A B C}}=\frac{P D}{A D}=\frac{12}{19} .
\end{array}
$$