4. Given point $P$ inside $\triangle A B C$, and satisfying $\overrightarrow{A P}=\frac{1}{3} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C}$, 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 $S_{1}: S_{2}: S_{3}=$ $\qquad$.
Answer $5: 4: 3$.
Analysis Because
$$
\overrightarrow{A P}=\frac{1}{3} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C}=\frac{1}{3}(\overrightarrow{P B}-\overrightarrow{P A})+\frac{1}{4}(\overrightarrow{P C}-\overrightarrow{P A})
$$

So
$$
5 \overrightarrow{P A}+4 \overrightarrow{P B}+3 \overrightarrow{P C}=\overrightarrow{0}
$$

Therefore
$$
S_{1}: S_{2}: S_{3}=5: 4: 3
$$