Denote the centres $(2, 1)$, $(10, 11)$ of the two circles $\Gamma_1, \Gamma_2$ as $T_1, T_2$, respectively.

If $l$ passes through $T_1$ or $T_2$, then $l$ bisects the circumference of $\Gamma_1$ or that of $\Gamma_2$, which yields two equal arcs. The possible slopes of $l$ at this point are $k_1 = k_{OT_1} = \frac{1}{2}$ or $k_2 = k_{OT_2} = \frac{11}{10}$.

If $l$ neither passes through $T_1$ nor $T_2$, then $\Gamma_1$ and $\Gamma_2$ are both divided into two arcs of unequal length by $l$. And since $\Gamma_1$ and $\Gamma_2$ are equal circles, the two arcs divided in $\Gamma_1$ are equal to the two arcs divided in $\Gamma_2$, respectively. This implies that $l$ is parallel to $T_1T_2$ or passes through its midpoint $M(6, 6)$. The possible slopes of $l$ at this point are $k_3 = k_{T_1T_2} = \frac{5}{4}$ or $k_4 = k_{OM} = 1$.

After checking, line $y = k_1x$ has no intersection with circle $\Gamma_2$, which does not fit the question; when $i = 2, 3, 4$, line $y = k_ix$ has two intersections with each of $\Gamma_1, \Gamma_2$, which is consistent with the question. Therefore, the sum of the slopes of all the lines $l$ satisfying the conditions is
$$k_2 + k_3 + k_4 = \frac{11}{10} + \frac{5}{4} + 1 = \frac{67}{20}. \quad \square$$