3. Given points $P, Q$ inside $\triangle A B C$, and
$$
\begin{array}{l}
\overrightarrow{P A}+2 \overrightarrow{P B}+3 \overrightarrow{P C} \\
=2 \overrightarrow{Q A}+3 \overrightarrow{Q B}+5 \overrightarrow{Q C}=\mathbf{0} . \\
\text { Then } \frac{|\overrightarrow{P Q}|}{|\overrightarrow{A B}|}=(\quad) .
\end{array}
$$
(A) $\frac{1}{30}$
(B) $\frac{1}{31}$
(C) $\frac{1}{32}$
(D) $\frac{1}{33}$
3. A.

From the given information,
$$
\begin{array}{l}
S_{\triangle B C P}: S_{\triangle C A P}: S_{\triangle A B P}=1: 2: 3, \\
S_{\triangle B C Q}: S_{\triangle C A Q}: S_{\triangle A B Q}=2: 3: 5 .
\end{array}
$$

Then, $S_{\triangle A B P}=S_{\triangle A B Q}=\frac{1}{2} S_{\triangle A B C}$.
Thus, $P Q \parallel A B$, and
$S_{\triangle B C P}=\frac{1}{6} S_{\triangle A B C}, S_{\triangle B C Q}=\frac{1}{5} S_{\triangle A B C}$.
Therefore, $\frac{|\overrightarrow{P Q}|}{|\overrightarrow{A B}|}=\frac{1}{5}-\frac{1}{6}=\frac{1}{30}$.