3. Let point $O$ be a point inside triangle $A B C$, and satisfy the relation
$$
\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=3 \overrightarrow{A B}+2 \overrightarrow{B C}+\overrightarrow{C A}
$$

Then $\frac{S_{\triangle A O B}+2 S_{\triangle B O C}+3 S_{\triangle C O A}}{S_{\triangle A B C}}=$ $\qquad$
Answer $\frac{11}{6}$.
Analysis
$$
\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=3 \overrightarrow{A B}+2 \overrightarrow{B C}+\overrightarrow{C A}
$$

can be transformed into
$$
3 \overrightarrow{O A}+\overrightarrow{O B}+2 \overrightarrow{O C}=\overrightarrow{0}, \quad(\overrightarrow{O A}+\overrightarrow{O B})+2(\overrightarrow{O A}+\overrightarrow{O C})=\overrightarrow{0}
$$

Let $M, N$ be the midpoints of $A B, A C$, respectively, then $\overrightarrow{O M}=-2 \overrightarrow{O N}$. Let the area of $\triangle A B C$ be $S$, by geometric relations we know
$$
S_{\triangle B O C}=\frac{1}{2} S, \quad S_{\triangle A O B}=\frac{1}{3} S, \quad S_{\triangle A O C}=\frac{1}{6} S
$$

Therefore,
$$
\frac{S_{\triangle A O B}+2 S_{\triangle B O C}+3 S_{\triangle C O A}}{S_{\triangle A B C}}=\frac{11}{6}
$$