3. As shown in Figure 2, in the equilateral $\triangle A B C$, side $A B$ is tangent to $\odot O$ at point $H$, and sides $B C$ and $C A$ intersect $\odot O$ at points $D, E, F, G$. Given that $A G=2, G F=6, F C$ $=1$. Then $D E=$ $\qquad$
3. $\sqrt{21}$.

By the secant-tangent theorem, we know $A H^{2}=A G \cdot A F=16$, which means $A H=4$.
Also, $A C=A G+G F+F C=9$, so, $A B=9$.
Therefore, $B H=5$.
Let $B D=x, C E=y$.
By the secant-tangent theorem, we know $B H^{2}=B D \cdot B E$, that is,
$$
25=x(9-y) \text {. }
$$

Also, $C E \cdot C D=C F \cdot C G$, that is,
$$
7=y(9-x) \text {. }
$$

Solving (1) and (2) together, we get $x=y+2$.
Thus, $y(7-y)=7$.
Solving this, we get $y=\frac{7 \pm \sqrt{21}}{2}$.
Since $x+y=2 y+2<9$, then $y<\frac{7}{2}$.
Therefore, $y=\frac{7-\sqrt{21}}{2}$, at this time, $x=\frac{11-\sqrt{21}}{2}$.
Thus, $D E=9-(x+y)=\sqrt{21}$.