Chord Circle Triad

Label the triangle $ABC$ , the chord $DE$ , $AL$ be the median of $\triangle ABC$ , the centers of the big circle, the middle green circle, and the right green circle be $O$ , $P$ , and $Q$ , $QM$ perpendicular to $DE$ , and the tangent point of the right green circle and the big circle be $N$ . Note that $O$ , $Q$ , and $N$ are colinear.

Let the radius of the green circles be $r$ and that of the big circle be $R$ . The median of $\triangle ABC$ , $AL = 12 \sqrt 3 \sin 60^\circ = 18$ . Note that center $P$ is the centroid of $\triangle ABC$ . Therefore $r = \dfrac 13 \cdot 18 = 6$ .

By Pythagorean theorem :

$\begin{aligned} OQ^2-OP^2 & = PQ^2 \\ (ON-NQ)^2 - (OA-AP)^2 & = (LC+CM)^2 \\ (ON-NQ)^2 - (OA-(AL-PL))^2 & = (LC + QM \cot 60^\circ)^2 \\ (R-r)^2 - (R-18+r)^2 & = \left(6\sqrt 3 + \frac r{\sqrt 3}\right) \\ (R-6)^2 - (R-12)^2 & = (6\sqrt 3 + 2 \sqrt 3)^2 \\ 12R - 108 & = 192 \\ \implies R & = \frac {300}{12} = \boxed{25} \end{aligned}$

From the given side, we know that the triangle height is 18 and green circle's radius is ⅓ of that, so r = 6. The distance between the centers of 2 closest green circles is double the horizontal distance from the center of middle green circle to its fencing triangle, so it is 2 x (1/2) x (2/3) x 12√3 = 8√3. Connect the centers of two closest green circles and the largest circle's to get a right triangle with sides 8√3, x and x + 6.
(x + 6)² = (8√3)² + x²
12x + 36 = 3 x 64
x = 16 - 3 = 13

Answer
= (x + 6) + 6
= 25