3. As shown in Figure 1, in quadrilateral $A B C D$, $\angle B=$
$$
\begin{array}{l}
135^{\circ}, \angle C=120^{\circ}, A B \\
=2 \sqrt{3}, B C=4-2 \sqrt{2}, \\
C D=4 \sqrt{2} \text {. Then the length of side } A D
\end{array}
$$

is ( ).
(A) $2 \sqrt{6}$
(B) $4 \sqrt{6}$
(C) $4+\sqrt{6}$
(D) $2+2 \sqrt{6}$
3. D.

As shown in Figure 8, draw $A E$ and $D F$ perpendicular to line $B C$, with the feet of the perpendiculars being $E$ and $F$ respectively.
From the given information, we have
$$
\begin{array}{c}
B E=A E=\sqrt{6}, \\
C F=2 \sqrt{2}, D F=2 \sqrt{6} .
\end{array}
$$

Thus, $E F=4+\sqrt{6}$.
Draw $A G \perp D F$, with the foot of the perpendicular being $G$. In the right triangle $\triangle A D G$, according to the Pythagorean theorem, we get
$$
\begin{array}{l}
A D=\sqrt{(4+\sqrt{6})^{2}+(\sqrt{6})^{2}} \\
=\sqrt{(2+\sqrt{24})^{2}}=2+2 \sqrt{6} .
\end{array}
$$