UPSC NDA II 2024 Mathematics Paper

Concept:The given equation represents a sphere. By converting it to standard form, we find the radius, then double it to get the diameter.Explanation:Start with the equation: $2x^2 + 2y^2 + 2z^2 + 3x + 3y + 3z - 6 = 0$.Divide the entire equation by 2 to simplify: $x^2 + y^2 + z^2 + \frac{3}{2}x + \frac{3}{2}y + \frac{3}{2}z - 3 = 0$.Complete the square for each variable: $x^2 + \frac{3}{2}x = \left(x + \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2$.Similarly for $y$ and $z$.Thus the equation becomes: $\left(x + \frac{3}{4}\right)^2 + \left(y + \frac{3}{4}\right)^2 + \left(z + \frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right)^2 - 3 = 0$.Calculate $3\left(\frac{3}{4}\right)^2 = \frac{27}{16}$.So: $\left(x + \frac{3}{4}\right)^2 + \left(y + \frac{3}{4}\right)^2 + \left(z + \frac{3}{4}\right)^2 = 3 + \frac{27}{16} = \frac{48}{16} + \frac{27}{16} = \frac{75}{16}$.The right side is the square of the radius: $R^2 = \frac{75}{16}$.Hence radius $R = \sqrt{\frac{75}{16}} = \frac{\sqrt{75}}{4} = \frac{5\sqrt{3}}{4}$.Diameter is twice the radius: $D = 2R = 2 \times \frac{5\sqrt{3}}{4} = \frac{5\sqrt{3}}{2}$.Thus the correct answer matches option B.Answer:$\frac{5\sqrt{3}}{2}$