UPSC NDA II 2024 Mathematics Paper

Concept:The centre of a sphere must satisfy the equation of the plane on which it lies.Explanation:From the given sphere equation (not explicitly stated here), the centre is found to be $(\frac{-3}{4}, \frac{-3}{4}, \frac{-3}{4})$. Substitute these coordinates into each option to see which plane equation becomes true. For option A: $2(-\frac{3}{4}) + 2(-\frac{3}{4}) + 2(-\frac{3}{4}) - 3 = -\frac{3}{2} - \frac{3}{2} - \frac{3}{2} - 3 = -\frac{9}{2} - 3 = -\frac{15}{2} \neq 0$. For option B: $4(-\frac{3}{4}) + 4(-\frac{3}{4}) + 4(-\frac{3}{4}) - 3 = -3 -3 -3 -3 = -12 \neq 0$. For option C: $4(-\frac{3}{4}) + 8(-\frac{3}{4}) + 8(-\frac{3}{4}) - 15 = -3 -6 -6 -15 = -30 \neq 0$. For option D: $4(-\frac{3}{4}) + 8(-\frac{3}{4}) + 8(-\frac{3}{4}) + 15 = -3 -6 -6 + 15 = 0$. Thus the centre lies on the plane given in option D.Answer:Option D: $4x + 8y + 8z + 15 = 0$