Let $O_{i}$ be the center and $r_{i}$ the radius of circle $\Gamma_{i}$ for each $i=1,2,3$. Let $P$ be an exceptional point, and let the three corresponding lines $A_{1} B_{1}, A_{2} B_{2}, A_{3} B_{3}$ concur at $Q$. Construct the circle with diameter $P Q$. Call the circle $\Gamma$, its center $O$ and its radius $r$. We now claim that all exceptional points lie on $\Gamma$.

Let $P O_{1}$ intersect $A_{1} B_{1}$ in $X_{1}$. As $P O_{1} \perp A_{1} B_{1}$, we see that $X_{1}$ lies on $\Gamma$. As $P A_{1}$ is a tangent to $\Gamma_{1}$, triangle $P A_{1} O_{1}$ is right-angled and similar to triangle $A_{1} X_{1} O_{1}$. It follows that
$$\frac{O_{1} X_{1}}{O_{1} A_{1}}=\frac{O_{1} A_{1}}{O_{1} P}, \quad \text{ i.e., } \quad O_{1} X_{1} \cdot O_{1} P=O_{1} A_{1}^{2}=r_{1}^{2}$$
On the other hand, $O_{1} X_{1} \cdot O_{1} P$ is also the power of $O_{1}$ with respect to $\Gamma$, so that
$$\begin{equation*} r_{1}^{2}=O_{1} X_{1} \cdot O_{1} P=\left(O_{1} O-r\right)\left(O_{1} O+r\right)=O_{1} O^{2}-r^{2} \tag{*} \end{equation*}$$
and hence
$$r^{2}=O O_{1}^{2}-r_{1}^{2}=\left(O O_{1}-r_{1}\right)\left(O O_{1}+r_{1}\right) .$$
Thus, $r^{2}$ is the power of $O$ with respect to $\Gamma_{1}$. By the same token, $r^{2}$ is also the power of $O$ with respect to $\Gamma_{2}$ and $\Gamma_{3}$. Hence $O$ must be the radical center of the three given circles. Since $r$, as the square root of the power of $O$ with respect to the three given circles, does not depend on $P$, it follows that all exceptional points lie on $\Gamma$.