3. In the convex quadrilateral $A B C D$, $\angle B=\angle D = 90^{\circ}$, $A D=D C$. If $S_{\text {quadrilateral } A B C D}=12$, then $A B+B C$ $=(\quad)$.
(A)6
(B) $4 \sqrt{3}$
(C) $5 \sqrt{2}$
(D) $3 \sqrt{6}$
3. B.

As shown in Figure 7, rotate $\triangle D A B$ to the position of $\triangle D C K$.

Since $A D=D C$, point $A$ must coincide with point $C$.
Since $\angle B=\angle D=90^{\circ}$,
it follows that $\angle A+\angle C=180^{\circ}$.
Therefore, points $B$, $C$, and $K$ are collinear.

At this point, $\triangle D B K$ is an isosceles right triangle, so

Thus, $D B=D K=2 \sqrt{6}$.
Hence, $A B+B C=B K=\sqrt{2} D B=4 \sqrt{3}$.