By Pitot's theorem, we have that $AD=7+2-3=6$. By Pythagorean theorem, we have that $AC=\sqrt{AB^2+BC^2}=\sqrt{13}$. Considering triangle $ACD$, we have that $AC^2=AD^2+CD^2-2\cdot AD\cdot CD\cdot \cos{\alpha}\implies \cos{\alpha}=\frac{6}{7}$, thus $\cos{\frac \alpha 2}=\sqrt{\frac{13}{14}}$. Let $CD$ be tangent to incircle at $E$ and let $BC$ be tangent to incircle at $F$. Thus, $DE=x=\sqrt{13}r$ and $CF=CE=y$. \[\left\{\begin{array}{l}x+y=7\\r+y=3\\x=\sqrt{13}r\end{array}\right.\]Thus, $\sqrt{13}r-r=4\implies r=\frac{4}{\sqrt{13}-1}=\frac{\sqrt{13}+1}{3}$.