3. In quadrilateral $A B C D$, it is known that $B C=8, C D=12$, $A D=10, \angle A=\angle B=60^{\circ}$. Then $A B=$
$$
3.9+\sqrt{141} \text {. }
$$

Extend $A D$ and $B C$ to intersect at point $P$.
Let $A B=x$. Then
$$
\begin{array}{l}
D P=x-10, \\
P C=x-8 .
\end{array}
$$

Draw $D H \perp$
$P C$ at point $H$.
In Rt $\triangle P H D$ and Rt $\triangle C H D$, we have respectively
$$
\begin{array}{l}
D H^{2}=D P^{2}-P H^{2}, D H^{2}=C D^{2}-H C^{2} \\
\Rightarrow D P^{2}-P H^{2}=C D^{2}-H C^{2} . \\
\text { Also } P H=\frac{P D}{2}=\frac{x-10}{2}, \\
H C=P C-P H=\frac{x-6}{2}, \\
\text { then }(x-10)^{2}-\left(\frac{x-10}{2}\right)^{2}=12^{2}-\left(\frac{x-6}{2}\right)^{2} \\
\Rightarrow A B=x=9+\sqrt{141} .
\end{array}
$$