Let the center of $\omega_{i}$ be $O_{i}$ for $i=1,2,3$ and let $O$ denote the center of $\Gamma$. Then $O, D$, and $O_{1}$ are collinear, as are $O, E$, and $O_{2}$. Denote by $F$ the point of tangency between $\Gamma$ and $\omega_{3}$; then $F, O$, and $O_{3}$ are collinear. Writing $r$ for the radius of $\Gamma$ we have $O O_{1}=r+2, O O_{2}=r+1, O O_{3}=3-r$. Now since $O_{1} O_{3}=1$ and $O_{3} O_{2}=2$, we apply Stewart's theorem: $$\begin{aligned} O O_{1}^{2} \cdot O_{2} O_{3}+O O_{2}^{2} \cdot O_{1} O_{3} & =O O_{3}^{2} \cdot O_{1} O_{2}+O_{1} O_{3} \cdot O_{3} O_{2} \cdot O_{1} O_{2} \\ 2(r+2)^{2}+(r+1)^{2} & =3(3-r)^{2}+1 \cdot 2 \cdot 3 \end{aligned}$$ We find $r=\frac{6}{7}$. Now the key observation is that the circumcircle of triangle $C D E$ is the incircle of triangle $O O_{1} O_{2}$. We easily compute the sides of $O O_{1} O_{2}$ to be $\frac{13}{7}, \frac{20}{7}$, and 3. By Heron's formula, the area of $O O_{1} O_{2}$ is $\frac{18}{7}$, but the semiperimeter is $\frac{27}{7}$, so the desired radius is $\frac{2}{3}$.