smc

Let $O=\Gamma$ be the center of the semicircle and $X=\Omega$ be the center of the circle. Applying the Extended Law of Sines to $△PQR,$ we find the radius of $\odot X:$ $$XP=\frac{QR}{2\cdot \sin \angle QPR}=\frac{3\sqrt{3}}{2\cdot \frac{\sqrt{3}}{2}}=3.$$ Alternatively, by the Inscribed Angle Theorem, $△QRX$ is a $30^{\circ }\text{-}30^{\circ }\text{-}120^{\circ }$ triangle with base $QR=3\sqrt{3}.$ Dividing $△QRX$ into two congruent $30^{\circ }\text{-}60^{\circ }\text{-}90^{\circ }$ triangles, we get that the radius of $\odot X$ is $XQ=XR=3$ by the side-length ratios. Let $M$ be the midpoint of $\overline{QR}.$ By the Perpendicular Chord Bisector Converse, we have $\overline{OM}\perp \overline{QR}$ and $\overline{XM}\perp \overline{QR}.$ Together, points $O,X,$ and $M$ must be collinear. By the SAS Congruence, we have $△QXM\cong △RXM,$ both of which are $30^{\circ }\text{-}60^{\circ }\text{-}90^{\circ }$ triangles. By the side-length ratios, we obtain $RM=\frac{3\sqrt{3}}{2},RX=3,$ and $XM=\frac{3}{2}.$ By the Pythagorean Theorem on right $△ORM,$ we get $OM=\frac{13}{2}$ and $OX=OM-XM=5.$ By the Pythagorean Theorem on right $△OXP,$ we get $OP=4.$ Let $C$ be the foot of the perpendicular from $P$ to $\overline{QR},$ and $D$ be the foot of the perpendicular from $X$ to $\overline{PC},$ as shown below: Clearly, quadrilateral $XDCM$ is a rectangle. Since $\angle XPD=\angle OXP$ by alternate interior angles, we have $△XPD\sim △OXP$ by the AA Similarity, with the ratio of similitude $\frac{XP}{OX}=\frac{3}{5}.$ Therefore, we get $PD=\frac{9}{5}$ and $PC=PD+DC=PD+XM=\frac{9}{5}+\frac{3}{2}=\frac{33}{10}.$ The area of $△PQR$ is $$\frac{1}{2}\cdot QR\cdot PC=\frac{1}{2}\cdot 3\sqrt{3}\cdot \frac{33}{10}=\frac{99\sqrt{3}}{20},$$ from which the answer is $99+3+20=\boxed{122}.$