Let $\{P_3\} = \Gamma_1 \cap \Gamma_2$ be the second point of intersection of the circles $\Gamma_1$ and $\Gamma_2$, and let $O_1, O_2, O_3$ be the centers of the circles $\Gamma_1, \Gamma_2$, and $\Gamma_3$, respectively. Denote by $O_4$ the intersection point of the common tangents to the circles $\Gamma_1$ and $\Gamma_2$.

$$\begin{align*} \widehat{P_1ZP_2} &= \frac{1}{2} (\widehat{AB} + \widehat{P_1P_2}) = \frac{1}{2} (180^\circ + \widehat{P_1O_3P_2}) \\ &= \frac{1}{2} (180^\circ + 180^\circ - \widehat{P_1O_2P_3} - \widehat{P_2O_1P_3}) = \\ &= \frac{1}{2} (180^\circ - \widehat{P_1O_2P_3}) + \frac{1}{2} (180^\circ - \widehat{P_2O_1P_3}) \\ &= \widehat{O_1P_3P_2} + \widehat{O_2P_3P_1} = 180^\circ - \widehat{P_1P_3P_2}, \end{align*}$$
so the quadrilateral $ZP_1P_3P_2$ is cyclic.

We will prove that $X, P_3$, and $Z$ are collinear. It suffices to show that $\widehat{XP_3O_1} = \widehat{ZP_3O_2}$. Since triangle $O_1XP_3$ is isosceles and $BP_2 \perp AX$ (because $AB$ is a diameter), we have:
$$\begin{align*} \widehat{XP_3O_1} &= \frac{1}{2} (180^\circ - \widehat{XO_1P_3}) = 90^\circ - \frac{1}{2} \widehat{XP_3} = \widehat{XP_2B} - \widehat{XP_2P_3} = \widehat{BP_2P_3} \\ &= \widehat{ZP_2P_3} = \frac{1}{2} \widehat{ZP_1P_3}. \end{align*}$$
Because $O_1O_2$ is tangent to the circumcircle of triangle $P_1P_2P_3$, it follows that $\widehat{P_1P_3O_2} = \frac{1}{2}\widehat{P_1P_3}$. Therefore, $\widehat{ZP_3O_2} = \widehat{ZP_3P_1} + \widehat{P_1P_3O_2} = \frac{1}{2}\widehat{ZP_1} + \frac{1}{2}\widehat{P_1P_3} = \frac{1}{2}\widehat{ZP_1P_3} = \widehat{XP_3O_1}$.
Hence, points $X, Z$, and $P_3$ are collinear. Similarly, one can show that $Y, Z$, and $P_3$ are collinear, and thus $X, Y$, and $Z$ are collinear as well.