3. Given that $M$ and $N$ are points on the hypotenuse $BC$ of an isosceles right $\triangle ABC$,
$$
AB = AC = 6 \sqrt{2}, BM = 3, \angle MAN = 45^{\circ}.
$$

Then $NC = (\quad)$.
(A) 3
(B) $\frac{7}{2}$
(C) 4
(D) $\frac{9}{2}$
3. C.

As shown in Figure 4, $\triangle C A N$ is rotated $90^{\circ}$ clockwise around point $A$ to $\triangle A B D$, and $M D$ is connected.
$$
\begin{array}{l}
\text { Then } \angle M B D=\angle A B M+\angle A B D \\
=45^{\circ}+45^{\circ}=90^{\circ}, \\
B D=C N, A D=A N, \\
\angle M A D=\angle B A C-\angle M A N \\
=45^{\circ}=\angle M A N .
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore, } \triangle A M N \cong \triangle A M D \Rightarrow M N=M D \text {. } \\
\text { It is easy to know, } B C=12, M C=9 . \\
\text { Thus, } M N+N C=9 . \\
\text { From } B D^{2}+B M^{2}=D M^{2} \\
\Rightarrow N C^{2}+3^{2}=M N^{2} .
\end{array}
$$

From equations (1) and (2), we solve to get $N C=4, M N=5$.