7. As shown in Fig. $3, \odot O$

is tangent to the hypotenuse $AB$ of the right triangle $\triangle ABC$ at point $D$, intersects the leg $AC$ at point $E$, and $DE \parallel BC$. Given $AE = 2\sqrt{2}$, $AC = 3\sqrt{2}$, and $BC = 6$. Then the radius of $\odot O$ is ( ).
(A) 3
(B) 4
(C) $4\sqrt{3}$
(D) $2\sqrt{3}$
7.D.

Notice $A B=\sqrt{(3 \sqrt{2})^{2}+6^{2}}=3 \sqrt{6}$.
From $D E / / B C \Rightarrow \triangle A E D \backsim \triangle A C B$
$$
\Rightarrow \frac{E D}{C B}=\frac{A E}{A C} \Rightarrow E D=6 \times \frac{2 \sqrt{2}}{3 \sqrt{2}}=4 \text {. }
$$

In Figure 3, draw $O F \perp D E$, with $F$ being the foot of the perpendicular, and connect $O D$. Then $F D=\frac{1}{2} E D=2$.
From $\triangle O F D \backsim \triangle B C A \Rightarrow \frac{O D}{B A}=\frac{F D}{C A}$.
Thus, the required radius $O D=3 \sqrt{6} \times \frac{2}{3 \sqrt{2}}=2 \sqrt{3}$.