3. As shown in Figure 4, in $\triangle A B C$, $\angle C=90^{\circ}, D$ is a point on side $B C$, $\angle A D C=3 \angle B A D, B D=9$, $D C=5$. Then $A B=$
3. 21 .

Construct the angle bisector $B E$ of $\angle A B C$, intersecting $A D$ at point $E$.
Then $\angle A B E=\angle D B E=\angle D A B$
$$
\begin{array}{l}
\Rightarrow A E=B E \Rightarrow \triangle D B E \backsim \triangle D A B \\
\Rightarrow \frac{D B}{D A}=\frac{D E}{D B}=\frac{B E}{A B} .
\end{array}
$$
$$
\text { Let } A B=m, A E=B E=x, A D=y, D E=y-x \text {. }
$$

Thus $\frac{9}{y}=\frac{x}{m}=\frac{y-x}{9}$
$$
\begin{array}{l}
\Rightarrow y^{2}=y x+81, y x=9 m \\
\Rightarrow y^{2}=9 m+81 .
\end{array}
$$

In the right triangle $\triangle A C D$, by the Pythagorean theorem,
$$
\begin{array}{l}
A C^{2}=m^{2}-14^{2}=9 m+81-5^{2} \\
\Rightarrow m^{2}-9 m-252=0 \\
\Rightarrow m=21 \Rightarrow A B=21 .
\end{array}
$$