Solution:

The circles have radius $\frac{1}{2}$ and centers at $\left(\frac{1}{2}, \frac{1}{2}\right),$ $\left(\frac{3}{2}, \frac{1}{2}\right),$ $\left(\frac{5}{2}, \frac{1}{2}\right),$ $\left(\frac{1}{2}, \frac{3}{2}\right),$ $\left(\frac{3}{2}, \frac{3}{2}\right),$ $\left(\frac{5}{2}, \frac{3}{2}\right),$ $\left(\frac{1}{2}, \frac{5}{2}\right),$ and $\left(\frac{3}{2}, \frac{5}{2}\right).$ The pair of circles tangent at $A = \left(1, \frac{1}{2}\right)$ is symmetric about $A,$ so any line through $A$ bisects that pair's area; similarly for the pair tangent at $B = \left(\frac{3}{2}, 2\right).$ The line $AB$ has slope $\frac{2 - 1/2}{3/2 - 1} = 3.$

Line $AB$ misses the remaining four circles entirely, and exactly two of them lie on each side of it, so it divides $\mathcal{R}$ into two regions of equal area. Sliding a slope-$3$ line strictly shifts area from one side to the other, so $\ell$ must be this line.

Its equation is $y - \frac{1}{2} = 3(x - 1),$ that is, $6x = 2y + 5.$ With $\gcd(6, 2, 5) = 1,$ the answer is $a^2 + b^2 + c^2 = 36 + 4 + 25 = 65.$