Here is my homework problem but slightly twisted to only prove the "if" not including "only if".
I have no clue about where to start and why is this even about homothety since I can't spot any parallel line or scaling and such. Even though I borrow the picture from Brialliant.org but there is no solution for the problem or any hint. I'm very thankful for any hint or short solution or anything as I spent a whole day getting no more information about the problem.
The key here is the following lemma:
Proof of the lemma: As $\omega$ is tangent internally to $\Omega$ at $T$, there is a homothety from $T$ that maps $\omega$ to $\Omega$. On the one hand, the image of the chord $AB$ is hence parallel to $AB$ and tangent to $\Omega$; obviously, the point of tangency must be the midpoint of the arc $AB$. On the other hand, the point of tangency is the image of $T$ and hence the point of intersection of $ST$ and $\Omega$. This proves the first part. For the second part, observe that $$\angle ASM=\angle ABM=\angle MAB=\angle MAT$$ and hence, the triangles $MTA$ and $MAS$ are similar. Consequently, $\frac{MA}{MT}=\frac{MS}{MA}$, which gives the desired result. QED.
Now to your problem: First assume that the three circles have a common tangent, which intersects $\Gamma$ at $P$ and $Q$. Let $M$ be the midpoint of the arc $PQ$ that does not contain $A,B,C$. On the one hand, since the power of $M$ wrt to any of the three circles $\Gamma_A, \Gamma_B, \Gamma_C$ is $MP^2=MQ^2$ by the lemma, $M$ is the radical center of these three circles. On the other hand, $\ell_{AB}$ is the radical axis of $\Gamma_A$ and $\Gamma_B$ and $\ell_{BC}$ is the radical axis of $\Gamma_B$ and $\Gamma_C$, so the point of intersection of $\ell_{AB}$ and $\ell_{BC}$ is the radical center of these three circles. In conclusion, $\ell_{AB}$ and $\ell_{BC}$ intersect at $M\in\Gamma$.
Now assume that $\ell_{AB}$ and $\ell_{BC}$ intersect at a point $X\in\Gamma$. Let $t_{AB}$ be the common exterior tangent of $\Gamma_A$ and $\Gamma_B$ with the property that $A,B$ and $X$ are on different sides of $t_{AB}$; define $P_{AB}$ and $Q_{AB}$ as the points of intersection of $t_{AB}$ and $\Gamma$. The objects $t_{BC}, P_{BC}, Q_{BC}$ are defined similarly. By the lemma, the midpoint $M_{AB}$ of the arc $P_{AB}Q_{AB}$ not containing $A,B$ has equal power $M_{AB}P_{AB}^2=M_{AB}Q_{AB}^2$ wrt to both $\Gamma_A$ and $\Gamma_B$; hence, $M_{AB}$ lies on the radical axis $\ell_{AB}$ of $\Gamma_A$ and $\Gamma_B$ from which we deduce $M_{AB}=X$. Similarly, $M_{BC}=X$, and so $X$ is the midpoint of both arcs $P_{AB}Q_{AB}$ and $P_{BC}Q_{BC}$. That means $\lbrace P_{AB}, Q_{AB}\rbrace=\lbrace P_{BC}, Q_{BC}\rbrace$ and hence $t_{AB}=t_{BC}$, which proves the claim.