24. As shown in Figure 3, the diameter $A B=14$ of the semicircle $\Gamma$, the circle $\Gamma^{\prime}$ is tangent to $A B$ at point $P$, and intersects the semicircle $\Gamma$ at points $Q$ and $R, Q R=$ $3 \sqrt{3}, \angle Q P R=60^{\circ}, S_{\triangle P Q R}=\frac{a \sqrt{b}}{c}\left(a 、 b 、 c \in \mathbf{Z}_{+}\right.$, $(a, c)=1$, $\sqrt{b}$ is the simplest radical form). Then the value of $a+b+c$ is $(\quad)$.
(A) 110
(B) 114
(C) 118
(D) 122
(E) 126
24. D.

Let $O, X$ be the centers of the semicircle $\Gamma$ and the circle $\Gamma^{\prime}$, respectively, with the radius of $\Gamma^{\prime}$ being $r$. Connect $O X$ and extend it to intersect $Q R$ at point $M$. Draw $P C \perp Q R$ at point $C$ from point $P$, and draw $X D \perp P C$ at point $D$ from point $X$. As shown in Figure 12.
In $\triangle P Q R$, by the Law of Sines, we have
$$
\begin{array}{l}
2 r=\frac{Q R}{\sin \angle Q P R}=\frac{3 \sqrt{3}}{\sin \frac{\pi}{3}}=6 \\
\Rightarrow r=P X=3 .
\end{array}
$$

It is easy to see that $O M \perp Q R, M R=\frac{1}{2} Q R=\frac{3 \sqrt{3}}{2}$.
Thus, $M X=C D=\sqrt{X R^{2}-R M^{2}}=\frac{3}{2}$,
$$
\begin{array}{l}
X O=O M-X M=\sqrt{O R^{2}-M R^{2}}-X M \\
=\sqrt{7^{2}-\left(\frac{3 \sqrt{3}}{2}\right)^{2}}-X M=\frac{13-3}{2}=5 .
\end{array}
$$

By $\triangle O P X \backsim \triangle X D P$
$$
\begin{aligned}
\Rightarrow D P=\frac{9}{5} \\
\Rightarrow P C=P D+D C=\frac{33}{10} .
\end{aligned}
$$

Thus, $S_{\triangle P Q R}=\frac{1}{2} Q R \cdot P C$
$$
=\frac{1}{2} \times 3 \sqrt{3} \times \frac{33}{10}=\frac{99 \sqrt{3}}{20} \text {. }
$$

Therefore, $a+b+c=20+99+3=122$.