4. Given point $P$ is inside $\triangle A B C$ and satisfies
$$
\overrightarrow{A P}=\frac{1}{3} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C} \text {, }
$$

and let the areas of $\triangle P B C, \triangle P C A, \triangle P A B$ be $S_{1}, S_{2}, S_{3}$, respectively. Then $S_{1}: S_{2}: S_{3}=$ $\qquad$
4. 5:4:3.

Notice that,
$$
\begin{array}{l}
\overrightarrow{A P}=\frac{1}{3} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C} \\
=\frac{1}{3}(\overrightarrow{P B}-\overrightarrow{P A})+\frac{1}{4}(\overrightarrow{P C}-\overrightarrow{P A}) .
\end{array}
$$

Then $5 \overrightarrow{P A}+4 \overrightarrow{P B}+3 \overrightarrow{P C}=\mathbf{0}$.
Thus $S_{1}: S_{2}: S_{3}=5: 4: 3$.