3. Let $G$ be the centroid of $\triangle A B C$, and $P Q$ passes through the centroid $G$, and satisfies
$$
\begin{array}{l}
\overrightarrow{C P}=m \overrightarrow{C A}, \overrightarrow{C Q}=n \overrightarrow{C B} . \\
\text { Then } \frac{1}{m}+\frac{1}{n}=
\end{array}
$$
3. 3 .

Connect $C G$ and extend it to intersect $A B$ at point $D$. Then $D$ is the midpoint of $A B$.
$$
\begin{array}{l}
\text { It is easy to see that, } \overrightarrow{C G}=\frac{2}{3} \overrightarrow{C D}=\frac{1}{3}(\overrightarrow{C A}+\overrightarrow{C B}) . \\
\text { Therefore, } \overrightarrow{P G}=\overrightarrow{C G}-\overrightarrow{C P}=\frac{1}{3}(\overrightarrow{C A}+\overrightarrow{C B})-m \overrightarrow{C A} \\
=\left(\frac{1}{3}-m\right) \overrightarrow{C A}+\frac{1}{3} \overrightarrow{C B}, \\
\overrightarrow{Q G}=\frac{1}{3} \overrightarrow{C A}+\left(\frac{1}{3}-n\right) \overrightarrow{C B} .
\end{array}
$$

Since $\overrightarrow{P G}$ and $\overrightarrow{Q G}$ are collinear, there exists $\lambda$ such that
$$
\begin{aligned}
& \overrightarrow{P G}=\lambda \overrightarrow{Q G} \\
\Rightarrow & \left(\frac{1}{3}-m\right) \overrightarrow{C A}+\frac{1}{3} \overrightarrow{C B} \\
= & \frac{\lambda}{3} \overrightarrow{C A}+\left(\frac{1}{3}-n\right) \lambda \overrightarrow{C B} \\
\Rightarrow & \frac{1}{3}-m=\frac{\lambda}{3}, \frac{1}{3}=\left(\frac{1}{3}-n\right) \lambda \\
\Rightarrow & \frac{1}{m}+\frac{1}{n}=3 .
\end{aligned}
$$