3. In $\triangle A B C$, it is known that $A B=12, A C=8, B C$ $=13, \angle A$'s angle bisector intersects the medians $B E$ and $C F$ at points $M$ and $N$, respectively. Let the centroid of $\triangle A B C$ be $G$. Then
$$
\frac{S_{\triangle G M N}}{S_{\triangle A B C}}=
$$
$\qquad$ .
3. $\frac{1}{168}$.

As shown in Figure 5, connect $E N$.
By the Angle Bisector Theorem, we have
$$
\begin{array}{l}
\frac{B M}{M E}=\frac{A B}{A E}=\frac{12}{4} \Rightarrow \frac{B G+G M}{G E-G M}=\frac{3}{1} \\
\Rightarrow \frac{B G+G M}{B G-2 G M}=\frac{3}{2} \Rightarrow \frac{3 G M}{3 B G}=\frac{3-2}{2 \times 3+2} \\
\Rightarrow \frac{G M}{B G}=\frac{1}{8} \Rightarrow \frac{G M}{B E}=\frac{1}{12} .
\end{array}
$$

Similarly, $\frac{G N}{C G}=\frac{1}{7}$.
$$
\begin{array}{l}
\text { Therefore, } \frac{S_{\triangle G M N}}{S_{\triangle A B C}}=\frac{S_{\triangle G M N}}{S_{\triangle G N E}} \cdot \frac{S_{\triangle G N E}}{S_{\triangle C E C}} \cdot \frac{S_{\triangle G E C}}{S_{\triangle B E C}} \cdot \frac{S_{\triangle B E C}}{S_{\triangle A B C}} \\
=\frac{G M}{B E} \cdot \frac{G N}{G C} \cdot \frac{C E}{C A}=\frac{1}{12} \times \frac{1}{7} \times \frac{1}{2}=\frac{1}{168} .
\end{array}
$$