Locus is $\Gamma_2$.

meysam1371 wrote: Locus is $\Gamma_2$. It's only a part of locus (because there are some special cases).

Consider external tangents of $\Gamma_i$'s meet at $O$. Let $O_i$ and $P$ be the center of $\Gamma_i$ and $\omega$. Assume $B$ is tangency point of $\Gamma_1$ and $\omega$. As in the the figure, $C$, $D$, $E$, $F$ and $G$ are other intersections of $OB$ with $\Gamma_i$'s and $l$. We name meeting point of internal tangents of $\Gamma_2$ and $\omega$, $C'$ and meeting point of internal tangents of $\Gamma_3$ and $\omega$, $F'$. Clearly $C \equiv C'$. So $O_2$ passes through $P$. We have: $$\mathcal{P}^{C}_{\Gamma_1} = \mathcal{P}^{C}_{\Gamma_2}, \quad \Rightarrow\quad \frac{CE}{CF} = \frac{CG}{CB}$$$$ O_1B || O_2G\quad \Rightarrow\quad \frac{CG}{CB} = \frac{CO_2}{CP}, \quad \Rightarrow\quad \frac{CE}{CF} = \frac{CO_2}{CP}$$So $$O_2E || FP.$$Also we know $O_2E || O_3F$. Thus $O_3$, $F$ and $P$ are colinear. On the other hand, $O$ is external homothetic center of $\Gamma_1$ and $\Gamma_3$, $B$ is internal homothetic center of $\Gamma_1$ and $\omega$, and, $F'$ is internal homothetic center of $\omega$ and $\Gamma_3$. So $F'$ lies on $OB$. Also $F'$ lies on $O_3P$. Thus $F \equiv F'$. We conclude that $F$ lies on $\Gamma_2$. $\blacksquare$ note: in the figure bellow, introduced point by $O$ in the solution, is shown by $A$. Edit: this solution is for the case that $\omega$ isn't tangent to $\Gamma_1$ and $l$ at same point. This special case is solved in two posts after this by JAnatolGT_00

JAnatolGT_00 wrote: meysam1371 wrote: Locus is $\Gamma_2$. It's only a part of locus (because there are some special cases). Would you specify these special cases please?

meysam1371 wrote: JAnatolGT_00 wrote: meysam1371 wrote: Locus is $\Gamma_2$. It's only a part of locus (because there are some special cases). Would you specify these special cases please? Observe that if $\omega$ touches $\Gamma_1$ at $A$, then the center of internal homothety between $\omega,\Gamma_3$ lies on segment $AB,$ and clearly all points of this segment can be obtained. Hence the answer is $\Gamma_2\cap AB\backslash \left\{ A,B\right\}.$

Yes, you are right. I will mention it in my solution. Thank you.