3. Given points $\mathrm{P}, \mathbf{Q}$ are inside $\triangle \mathbf{A B C}$, and $\overrightarrow{P A}+2 \overrightarrow{P B}+3 \overrightarrow{P C}=2 \overrightarrow{Q A}+3 \overrightarrow{P B}+5 \overrightarrow{P C}=\overrightarrow{0}$, then $\frac{|\overrightarrow{P Q}|}{\mid \overrightarrow{A B}}$ equals
( ) .
A. $\frac{1}{30}$
B. $\frac{1}{31}$
c. $\frac{1}{32}$
D. $\frac{1}{33}$
【Answer】A
【Analysis】【Detailed Explanation】
From the given, we know $S_{\triangle B C P}: S_{\triangle C A P}: S_{\triangle A B P}=1: 2: 3, S_{\triangle B C Q}: S_{\triangle C A Q}: S_{\triangle A B Q}=2: 3: 5$,
$S_{\triangle A B P}=S_{\triangle A B Q}=\frac{S_{\triangle A B C}}{2}$, so $P Q / / A B$.
Also, $S_{\triangle B C P}=\frac{S_{\triangle A B C}}{6}, S_{\triangle B C Q}=\frac{S_{\triangle A B C}}{5}$,
$\frac{|\overrightarrow{P Q}|}{|\overrightarrow{A B}|}=\frac{1}{5}-\frac{1}{6}=\frac{1}{30}$.
Therefore, the answer is: $\mathrm{A}$