UPSC NDA I 2023 Mathematics Solved Paper

Concept:For any point $P$ on a sphere, if $A$ and $B$ are the endpoints of a diameter, then $\angle APB = 90^{\circ}$, making triangle $PAB$ right-angled at $P$.Explanation:Given $A(1, -1, 2)$ and $B(2, 1, -1)$ are the endpoints of a diameter of the sphere.The length of the diameter $AB$ is:$AB = \sqrt{(1-2)^2 + (-1-1)^2 + (2+1)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$Since $P$ lies on the sphere, $\angle APB = 90^{\circ}$ (angle in a semicircle).Thus $\triangle PAB$ is right-angled at $P$, so by the Pythagorean theorem:$PA^2 + PB^2 = AB^2$Substituting $AB = \sqrt{14}$, we get:$PA^2 + PB^2 = (\sqrt{14})^2 = 14$Answer:$14$ (Option B).