3. In $\triangle A B C$, $G$ is the centroid, and $I$ is the intersection of the angle bisectors of $\angle B$ and $\angle C$. If $I G / / B C$, and $B C=5$, then $A B+A C$ $=$ . $\qquad$
3. Connect $A G, B G$ and $C G$, and extend $A G$ to intersect $B C$ at $M$, then $S_{\triangle G B C}: S_{\triangle A B C}=G M: A M=1: 3$.

Connect $A I$, then by $I G$ $/ / B C$, we know
$$
\begin{array}{l}
S_{\triangle I B C}=S_{\triangle G B C}=\frac{S_{\triangle A B C}}{3}, \\
\therefore S_{\triangle I A B}+S_{\triangle I A C}=2 S_{\triangle I B C} .
\end{array}
$$
$\because I$ is the intersection of the angle bisectors of $\angle B$ and $\angle C$ in $\triangle A B C$,
$\therefore I$ is equidistant from the three sides of $\triangle A B C$.
Thus, $A B+A C=2 B C=10$.