The more the merrier ....

Letting $P, Q, R, S$ represent the centers of the respective circles, we have that $\Delta PRS$ is equilateral with side lengths $2$ . Now $\Delta PRQ$ is isosceles with $\angle PRQ = 120^{\circ}$ , and so $\angle PQR = 30^{\circ}$ . Thus $\Delta PQS$ is a $30/60/90$ right triangle with side lengths $PS = 2, QS = 4$ and $PQ = 2\sqrt{3}$ , giving us a perimeter for $\Delta PQS$ of $6 + 2\sqrt{3}$ .

Now as the $4$ circles have the same radius we can conclude that $\Delta ABC$ is a $30/60/90$ triangle as well. So if we can find the ratio of $AB$ to $PS$ we can multiply this ratio by the perimeter of $\Delta PQS$ to get the perimeter of $\Delta ABC$ .

Now draw perpendiculars from $P$ and $S$ to $AB$ intersecting at the respective points $M,N$ . Then since $\angle PAM = 45^{\circ}$ and $\angle SCN = 30^{\circ}$ we have that

$AB = AM + MN + NC = 1 + 2 + \sqrt{3} = 3 + \sqrt{3}$ .

The perimeter of $\Delta ABC$ is then

$\dfrac{(3 + \sqrt{3})}{2} * (6 + 2\sqrt{3}) = 12 + 6\sqrt{3}$ , and so

$a + b + c = 12 + 6 + 3 = \boxed{21}$ .

$PRS~ is~ a~ 2-2-2 ~\Delta. ~ Since ~ R ~ and ~ S ~ are ~ given, ~ P(3, 1+\sqrt3).\\ \text{Let M and N be the points of tangency. So MPSN is a rectangle. Implies MN=PS=2}\\ AC||PS ~ \therefore ~ slope ~ of ~ AC= ~ - ~ \sqrt3. ~ \implies ~ BCA=60^o.\\ AB||PQ ~ \therefore ~ slope ~ of ~ AB= ~ \dfrac {\sqrt3} 3. ~ \implies ~ ABC=30^o.\\ \implies ~ ABC ~ is ~ a ~ rt. ~ \angle ed ~ \Delta, ~ rt. ~ \angle ed ~ at ~ A.\\ \therefore ~ Tangent ~length ~ from ~ AM ~ =1 \because ~ it ~ is ~ a ~ side ~ of ~ a ~ square..\\ Tangent ~ length ~ from ~ C=\dfrac 1 {Cot\frac{60} 2}=\sqrt 3.\\ So ~ AC=AM+MN+NC=1+2+\sqrt3=3+\sqrt 3. \\ Perimeter ~ of ~ 30-60-90 ~ \Delta ~ ABC=(1+\sqrt3+2)*AC=\left (3+\sqrt3 \right)^2\\ =12+6*\sqrt3. ~~a+b+c=\boxed{21}$