Answer: D

Solution:

In circle $\Omega,$ the chord $QR$ subtends the inscribed angle $\angle QPR = 60^\circ,$ so $QR = 2r\sin 60^\circ,$ giving $3\sqrt3 = r\sqrt3,$ hence $r = 3.$

Place $A = (-7, 0),$ $B = (7, 0),$ with $\Gamma: x^2 + y^2 = 49$ (upper half). Since $\Omega$ is tangent to $AB$ at $P = (p, 0),$ its center is $(p, 3).$ Subtracting the two circle equations gives the line $QR,$ and the distance from the center $(p,3)$ to $QR$ must equal $r\cos 60^\circ = \tfrac32.$ This yields $(p^2 - 31)^2 = 9p^2 + 81,$ so $p^2 = 16$ (the root $p^2 = 55$ places $P$ outside $AB$).

With $p^2 = 16,$ the distance from $P$ to line $QR$ is $\dfrac{49 - p^2}{\sqrt{4p^2 + 36}} = \dfrac{33}{10}.$ Thus $$\begin{aligned} [\triangle PQR] &= \tfrac12\cdot QR\cdot d \\ &= \tfrac12\cdot 3\sqrt3\cdot\tfrac{33}{10} \\ &= \frac{99\sqrt3}{20}. \end{aligned}$$ So $a = 99,$ $b = 3,$ $c = 20,$ and $a + b + c = 122.$

Thus, the correct answer is D.