By Desargue's Theorem, we just have to prove that $ PP_1, QQ_1, RR_1$ are concurrent . So we have to prove that meet of $ PP_1$ and $ QQ_1$ lies on $ RR_1$. $ RR_1$ is the radical axis of $ \Gamma_{1}$ and $ \Gamma_{2}$. So we have to prove that $ X$ lies on the the radical axis of $ \Gamma_{1}$ and $ \Gamma_{2}$, where $ X$ is the intersection of $ PP_1$ and $ QQ_1$. That is, $ X$ has equal power with respect to the two circles $ \Gamma_{1}$ and $ \Gamma_{2}$. We prove this by the definition of power of points, or in other words, quadrilateral $ PQQ_1P_1$ is cyclic. Using powers of $ O_1, O_2,O_3$, we can easily get the cyclic condition. Hence done.

Since $ P$ and $ Q$ are exsimilicenters of $ \Gamma_1 \sim \Gamma$ and $ \Gamma_2 \sim \Gamma,$ it follows that $ O_1 \equiv PQ \cap P_1Q_1$ is the exsimilicenter of $ \Gamma_1,\Gamma_2.$ $ O_1$ is also center of the inversion with power $ O_1R^2$ taking $ \Gamma_1$ and $ \Gamma_2$ into each other, therefore the pairs $ P_1,Q_1$ and $ P,Q$ are inverse points $ \Longrightarrow$ $ PQQ_1P_1$ is cyclic. Let $ \Gamma_3$ denote its circumcircle. $ RR_1,PP_1,QQ_1$ concur at the radical center $ S$ of $\Gamma_1,$ $\Gamma_2,$ $\Gamma_3.$ By Desargues theorem, the triangles $\triangle PRQ$ and $\triangle P_1R_1Q_1$ are perspective through $ S$ $ \Longrightarrow$ Intersections $ O_1 \equiv PQ \cap P_1Q_1,$ $O_2 \equiv PR \cap P_1R_1,$ $O_3 \equiv RQ \cap R_1Q_1$ are collinear, as desired.

How to prove PQQ1P1 by powerpoibt since your explanation sir Agr_94_math is not very clear we cant saya tge intersection is at the radical axis,isnt it???