3. As shown in Figure 1, in the right trapezoid $A B C D$, the bases $A B=13, C D=8, A D \perp A B$ and $A D$ $=$ 12. Then the distance from $A$ to $B C$ is ( ).
(A) 12
(B) 13
(C) $\frac{12 \times 21}{13}$
(D) 10.5
3. (A).

Draw $C E \perp A B$ at $E$, then $A E C D$ is a rectangle.
Therefore, $A E=D C=8, C E=A D=12$.
Thus, $B E=A B-A E=5, B C=\sqrt{C E^{2}+B E^{2}}=$ 13, which means $\triangle A B C$ is an isosceles triangle, with the heights from the two equal sides being equal, so $A F$ $=C E=12$.