21. In tetrahedron $ABCD$, it is known that $AB=5, AC=3, BC=4, BD=4, AD=3, CD=\frac{12 \sqrt{2}}{5}$. Then the volume of tetrahedron $ABCD$ is $(\quad$.
(A) $3 \sqrt{2}$
(B) $2 \sqrt{5}$
(C) $3 \sqrt{3}$
(D) $\frac{24}{5}$
(E) $\frac{24 \sqrt{2}}{5}$
21. D.

As shown in Figure 6, in the tetrahedron $ABCD$, take the midpoint $E$ of edge $CD$, and connect $AE$ and $BE$.
From the problem, we know that
$AE \perp CD$, $BE \perp CD$.
Thus, $CE \perp$ plane $ABE$.
Given $CD=\frac{12 \sqrt{2}}{5}$, we get $CE=DE=\frac{6 \sqrt{2}}{5}$.
Therefore, $AE=\frac{\sqrt{153}}{5}, BE=\frac{\sqrt{328}}{5}$.
Then, $\cos \angle ABE=\frac{AB^{2}+BE^{2}-AE^{2}}{2 AB \cdot BE}$
$=\frac{25+\frac{328}{25}-\frac{153}{25}}{2 \times 5 \times \frac{\sqrt{328}}{5}}=\frac{32}{2 \sqrt{328}}=\frac{4 \sqrt{82}}{41}$
$\Rightarrow \sin \angle ABE=\frac{3 \sqrt{41}}{41}$.
Thus, $S_{\triangle ABE}=\frac{1}{2} AB \cdot BE \sin \angle ABE$
$$
=\frac{1}{2} \times 5 \times \frac{\sqrt{328}}{5} \times \frac{3 \sqrt{41}}{41}=3 \sqrt{2} \text {. }
$$

Therefore, the volume of the tetrahedron $ABCD$
$$
V=\frac{1}{3} S_{\triangle ABE} \cdot CD=\frac{1}{3} \times 3 \sqrt{2} \times \frac{12 \sqrt{2}}{5}=\frac{24}{5} .
$$