Question:

Step 1: Identify point $P$. The point $P$ with $PA:PB:PC = \cosec(A/2):\cosec(B/2):\cosec(C/2)$ is the incenter $I$ of $\triangle ABC$. For the incenter, $IA = r/\sin(A/2)$, $IB = r/\sin(B/2)$, $IC = r/\sin(C/2)$, so $IA:IB:IC = \cosec(A/2):\cosec(B/2):\cosec(C/2)$. Step 2: Compute basic quantities. $a = BC = 9$, $b = CA = 8$, $c = AB = 7$. $$s = \frac{9+8+7}{2} = 12$$ $$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12\cdot3\cdot4\cdot5} = \sqrt{720} = 12\sqrt{5}$$ $$r = \frac{\Delta}{s} = \frac{12\sqrt{5}}{12} = \sqrt{5}$$ Step 3: Compute $\cos A$, $\cos B$, $\cos C$. $$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{64+49-81}{2\cdot8\cdot7} = \frac{32}{112} = \frac{2}{7}$$ $$\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{81+49-64}{2\cdot9\cdot7} = \frac{66}{126} = \frac{11}{21}$$ $$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{81+64-49}{2\cdot9\cdot8} = \frac{96}{144} = \frac{2}{3}$$ $$\cos A+\cos B+\cos C = \frac{2}{7}+\frac{11}{21}+\frac{2}{3} = \frac{6}{21}+\frac{11}{21}+\frac{14}{21} = \frac{31}{21}$$ Option (b) is correct. Step 4: Compute $PA^2+PB^2+PC^2$ where $P$ is the incenter. $$IA^2 = \frac{r^2}{\sin^2(A/2)}, \quad \sin^2\frac{A}{2} = \frac{(s-b)(s-c)}{bc} = \frac{4\cdot5}{56} = \frac{20}{56} = \frac{5}{14}$$ $$IA^2 = \frac{5}{5/14} = 14$$ $$\sin^2\frac{B}{2} = \frac{(s-a)(s-c)}{ac} = \frac{3\cdot5}{63} = \frac{15}{63} = \frac{5}{21}$$ $$IB^2 = \frac{5}{5/21} = 21$$ $$\sin^2\frac{C}{2} = \frac{(s-a)(s-b)}{ab} = \frac{3\cdot4}{72} = \frac{12}{72} = \frac{1}{6}$$ $$IC^2 = \frac{5}{1/6} = 30$$ $$PA^2+PB^2+PC^2 = 14+21+30 = 65$$ Option (c) is correct. Step 5: Identify point $Q$. $AQ=BQ=CQ$ means $Q$ is the circumcenter of $\triangle ABC$. $$R = \frac{abc}{4\Delta} = \frac{9\cdot8\cdot7}{4\cdot12\sqrt{5}} = \frac{504}{48\sqrt{5}} = \frac{21}{2\sqrt{5}} = \frac{21\sqrt{5}}{10}$$ $$AQ = R = \frac{21\sqrt{5}}{10}$$ Option (d) is correct.