Let $O_1, O_2, O_3$ be respectively the centers of the circles $\Gamma_1, \Gamma_2, \Gamma_3$, and $R_1, R_2, R_3$ be their radii.
Let $\Gamma$ be the circumscribed circle of the triangle $O_1O_2O_3$ (see Fig. 1), and let $\Gamma$ touch the lines $O_3O_1, O_3O_2, O_1O_2$ at points $K_1, K_2, K_3$ respectively. Then
$$O_1K_1 = O_1K_3, \quad O_2K_2 = O_2K_3 \implies O_1K_1 + O_2K_2 = O_1O_2 = R_1 + R_2. \quad (1)$$

Moreover
$$(R_3 - R_1) + O_1K_1 = O_3O_1 + O_1K_1 = O_3K_1 = O_3K_2 = O_3O_2 + O_2K_2 = (R_3 - R_2) + O_2K_2.$$

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Therefore, if $O_1K_1 > R_1$ ($O_1K_1 < R_1$), then $O_2K_2 = O_3K_2 - (R_3 - R_2) = (R_3 - R_1) + O_1K_1 - (R_3 - R_2) = R_2 + (O_1K_1 - R_1) > R_2$, ($O_2K_2 < R_2$), which gives $O_1K_1 + O_2K_2 > R_1 + R_2$ ($O_1K_1 + O_2K_2 < R_1 + R_2$), contrary to (1). Therefore, $O_1K = R_1$, $O_2K_2 = R_2$. So $K_1, K_2, K_3$ coincide with $M_1, M_2, M_3$ respectively. It means that $\Gamma$ is circumcircle of the triangle $M_1M_2M_3$ (see Fig. 2). So $SM_1 \perp O_3O_1$ (as the radius of the inscribed circle), thus $SM_1$ is the tangent to $\Gamma_3$.