3. As shown in Figure 3, in the right triangle $\triangle ABC$, $BC=3, AC=$ $4, AB=5$, and its incircle is $\odot O$. Through $OA, OB, OC$ and the intersection points $M, N, K$ with $\odot O$, draw the tangents to $\odot O$, which intersect the three sides of $\triangle ABC$ at $A_{1}, A_{2}, B_{1}, B_{2}, C_{1}, C_{2}$. Then the area of the hexagon $A_{1} A_{2} B_{1} B_{2} C_{1} C_{2}$ is
$3.2 \sqrt{2}+\sqrt{5}+\frac{2 \sqrt{10}}{3}-\frac{11}{3}$.
As shown in Figure 7, let the radius of $\odot O$ be $r$, and $\odot O$ is tangent to $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$, respectively. Connect $OD$, $OE$, and $OF$. Then
$$
\begin{array}{c}
S_{\triangle ABC}=6 . \\
\text { By } \frac{1}{2} r(AB+BC \\
+CA)=S_{\triangle ABC}, \text { we get } \\
r=1 .
\end{array}
$$

Thus, $CD=CE=1$,
$$
\begin{array}{l}
BD=BF=2, AE=AF=3, \\
OA=\sqrt{OE^{2}+AE^{2}}=\sqrt{10}, \\
OB=\sqrt{OD^{2}+BD^{2}}=\sqrt{5}, OC=\sqrt{2}, \\
AM=\sqrt{10}-1, BN=\sqrt{5}-1, CK=\sqrt{2}-1, \\
S_{\triangle AOE}=\frac{3}{2}, S_{\triangle BOD}=1, S_{\triangle COD}=\frac{1}{2} . \\
\frac{S_{\triangle MMA_{1}}}{S_{\triangle AOE}}=\left(\frac{AM}{AE}\right)^{2}=\left(\frac{\sqrt{10}-1}{3}\right)^{2}=\frac{11-2 \sqrt{10}}{9},
\end{array}
$$

Thus, $S_{\triangle MMA_{1}}=\frac{11-2 \sqrt{10}}{6}, S_{\triangle A_{1}A_{2}}=\frac{11-2 \sqrt{10}}{3}$;
$$
\frac{S_{\triangle BVB_{1}}}{S_{\triangle BOD}}=\left(\frac{BN}{BF}\right)^{2}=\left(\frac{\sqrt{5}-1}{2}\right)^{2}=\frac{6-2 \sqrt{5}}{4} \text {, }
$$

Thus, $S_{\triangle BVB_{1}}=\frac{6-2 \sqrt{5}}{4}, S_{\triangle BB_{1}B_{2}}=\frac{6-2 \sqrt{5}}{2}$;
$$
\frac{S_{\triangle CKC_{1}}}{S_{\triangle COD}}=\left(\frac{CK}{CD}\right)^{2}=\left(\frac{\sqrt{2}-1}{1}\right)^{2}=3-2 \sqrt{2} \text {, }
$$

Thus, $S_{\triangle CKC_{1}}=\frac{3-2 \sqrt{2}}{2}, S_{\triangle CC_{1}C_{2}}=3-2 \sqrt{2}$.
Therefore, $S_{\text {hexagon } A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}}$
$$
\begin{array}{l}
=S_{\triangle ABC}-S_{\triangle AA_{1}A_{2}}-S_{\triangle BB_{1}B_{2}}-S_{\triangle C_{1}C_{2}} \\
=2 \sqrt{2}+\sqrt{5}+\frac{2 \sqrt{10}}{3}-\frac{11}{3} .
\end{array}
$$