UPSC NDA II 2025 Mathematics Paper

Concept:The sphere passes through three points. Its center must be equidistant from them. For the smallest possible radius, the center is the circumcenter of the triangle formed by the points.Explanation:Let the centre be $O(x,y,z)$.Equate distances from $O$ to $(1,0,0)$ and $(0,1,0)$:$$(x-1)^2 + y^2 + z^2 = x^2 + (y-1)^2 + z^2$$Simplify to get $x = y$.Equate distances from $O$ to $(0,1,0)$ and $(0,0,1)$:$$x^2 + (y-1)^2 + z^2 = x^2 + y^2 + (z-1)^2$$Simplify to get $y = z$. Thus $x = y = z$.For the smallest sphere passing through the three points, the centre is the circumcenter of the triangle. The circumcenter of an equilateral triangle is its centroid, so:$$x = y = z = \frac{1}{3}$$Now compute the radius $r$ as the distance from $O$ to $(1,0,0)$:$$r = \sqrt{\left(1-\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2}$$$$= \sqrt{\frac{4}{9} + \frac{1}{9} + \frac{1}{9}} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}}$$Answer:$\sqrt{\frac{2}{3}}$ (Option B)