4. Given point $P$ 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 $\triangle P B C$, $\triangle P C A$, and $\triangle P A B$ be $S_{1}$, $S_{2}$, and $S_{3}$, 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 $5 \overrightarrow{P A}+4 \overrightarrow{P B}+3 \overrightarrow{P C}=0$.
Thus $S_{1}: S_{2}: S_{3}=5: 4: 3$.