NDA Mathematics Question (2025) — Solution

Step-by-step solution

The sphere S passes through the points A(1, 0, 0), B(0, 1, 0), and C(0, 0, 1). The equation of the plane containing these three points is given by the intercept form: x 1 + y 1 + z 1 = 1 x + y + z - 1 = 0 Any sphere passing through A, B, and C will intersect this plane in a circle, which is the circumcircle of ABC. For the sphere to have the smallest possible radius, this circumcircle must be a great circle of the sphere. This means the centre of the sphere must lie exactly on the plane containing the points A, B, and C. Thus, the centre of the sphere S lies on the plane x + y + z - 1 = 0. Alternatively, ABC is an equilateral triangle. Its circumcentre is the same as its centroid, which is ( 1+0+0 3 , 0+1+0 3 , 0+0+1 3 ) = ( 1 3 , 1 3 , 1 3 ). Substituting ( 1 3 , 1 3 , 1 3 ) into the given options, only x + y + z - 1 = 0 is satisfied. Answer: x + y + z - 1 = 0