(4) Given point $O$ is inside $\triangle A B C$, and $3 \overrightarrow{A B}+2 \overrightarrow{B C}+\overrightarrow{C A}=4 \overrightarrow{A O}$, let the area of $\triangle A B C$ be $S_{1}$, and the area of $\triangle O B C$ be $S_{2}$. Then the value of $\frac{S_{1}}{S_{2}}$ is
(4) 2 Hint: From $3 \overrightarrow{A B}+2 \overrightarrow{B C}+\overrightarrow{C A}=4 \overrightarrow{A O}$, we know that
$$
3 \overrightarrow{A B}+2(\overrightarrow{A C}-\overrightarrow{A B})-\overrightarrow{A C}=4 \overrightarrow{A O}.
$$

Therefore, $\overrightarrow{A B}+\overrightarrow{A C}=4 \overrightarrow{A O}$. Taking the midpoint $D$ of $B C$, we have $2 \overrightarrow{A D}=4 \overrightarrow{A O}$, $\overrightarrow{A D}=2 \overrightarrow{A O}$.
Thus, $A, O, D$ are collinear, and $O$ is the midpoint of $A D$. Therefore, $\frac{S_{1}}{S_{2}}=\frac{A D}{O D}=2$.