6. As shown in Figure 2, $C D$ is the altitude on the hypotenuse $A B$ of Rt $\triangle A B C$, $I_{1} 、 I_{2} 、 I_{3} 、$ are the incenter of $\triangle A B C 、 \triangle A C D$ 、 $\triangle B C D$ respectively, $A C = 20, B C = 15$. Then the area of $\triangle I_{1} I_{2} I_{3}$ is ( ).
(A) 4
(B) 4.5
(C) 5
(D) 5.5
6.C.

As shown in Figure 6, connect $A I_{1}$, 
$B I_{1}$, $I_{2} D$, $I_{3} D$. Draw $I_{1} G \perp A B$ at point $G$, 
$I_{2} E \perp A B$ at point $E$, and $I_{3} F \perp A B$ 
at point $F$. In the right triangle $\triangle A B C$, it is easy to see that
$$
A B=\sqrt{A C^{2}+B C^{2}}=\sqrt{20^{2}+15^{2}}=25 \text {. }
$$

Since $C D \perp A B$, by the projection theorem, we have
$$
A D=\frac{A C^{2}}{A B}=\frac{20^{2}}{25}=16 \text {. }
$$

Thus, $B D=A B-A D=9$,
$$
C D=\sqrt{A C^{2}-A D^{2}}=\sqrt{20^{2}-16^{2}}=12 \text {. }
$$

Since $I_{1}$, $I_{2}$, and $I_{3}$ are the incenter of $\triangle A B C$, $\triangle A C D$, and $\triangle B C D$ respectively, $I_{2}$ lies on $A I_{1}$, $I_{3}$ lies on $B I_{1}$, and $I_{1} G$, $I_{2} E$, $I_{3} F$ are the inradii of $\triangle A B C$, $\triangle A C D$, and $\triangle B C D$ respectively. $D I_{2}$ and $D I_{3}$ are the angle bisectors of $\angle A D C$ and $\angle B D C$ respectively. Therefore,
$$
\begin{array}{l}
\angle A D I_{2}=\angle I_{2} D C=45^{\circ}, \\
\angle B D I_{3}=\angle C D I_{3}=45^{\circ} \\
\Rightarrow \angle I_{2} D I_{3}=\angle I_{2} D C+\angle C D I_{3}=90^{\circ} \\
\Rightarrow I_{2} D \perp I_{3} D .
\end{array}
$$

Since $I_{1} G$ is the inradius of the right triangle $\triangle A B C$, we have
$$
I_{1} G=\frac{1}{2}(A C+B C-A B)=5 \text {. }
$$

Similarly, $I_{2} E=4$, $I_{3} F=3$.
Thus, $I_{2} D=\frac{I_{2} E}{\sin \angle A D I_{2}}=\frac{4}{\sin 45^{\circ}}=4 \sqrt{2}$.
Similarly, $I_{3} D=3 \sqrt{2}$.
Therefore, $S_{\triangle I_{1} I_{2} I_{3}}$
$$
\begin{aligned}
= & S_{\triangle A B B_{1}}-S_{\triangle \triangle I_{2}}-S_{\triangle B D I_{3}}-S_{\triangle D I_{2} I_{3}} \\
= & \frac{1}{2} A B \cdot I_{1} G-\frac{1}{2} A D \cdot I_{2} E- \\
& \frac{1}{2} B D \cdot I_{3} F-\frac{1}{2} I_{2} D \cdot I_{3} D \\
= & 5 .
\end{aligned}
$$