Solution:

By Heron's formula we have $[A B C] = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84$.

Let $D$ be the foot of the altitude from $A$ to $B C$; then $A D = 2 \cdot \frac{84}{14} = 12$.

Notice that because $\angle B F C = \angle B E C$, $B F E C$ is cyclic, so $\angle A F E = 90^\circ - \angle E F C = 90^\circ - \angle E B C = \angle C$.

Therefore, we have $\triangle A E F \sim \triangle A B C$, so $\frac{A G}{A D} = \frac{A E}{A B}$.

$\frac{1}{2} (B E)(A C) = 84 \Longrightarrow B E = \frac{56}{5}$

$A E = \sqrt{13^2 - \left(\frac{56}{5}\right)^2} = \sqrt{\frac{65^2 - 56^2}{5^2}} = \frac{33}{5}$.

Then $A G = A D \cdot \frac{A E}{A B} = 12 \cdot \frac{33 / 5}{13} = \frac{396}{65}$.