UPSC NDA I 2024 Mathematics Paper

Concept:Use the factorization $x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]$ and the property of an equilateral triangle.Explanation:Given $\sin^3 A + \sin^3 B + \sin^3 C = 3 \sin A \sin B \sin C$.Rewrite as $\sin^3 A+\sin^3 B+\sin^3 C - 3\sin A\sin B\sin C = 0$.Apply the identity with $x=\sin A$, $y=\sin B$, $z=\sin C$:$\frac{1}{2}(\sin A+\sin B+\sin C)[(\sin A-\sin B)^2+(\sin B-\sin C)^2+(\sin C-\sin A)^2] = 0$.In a triangle, all angles lie in $(0,\pi)$, so $\sin A,\sin B,\sin C > 0$.Hence $\sin A+\sin B+\sin C > 0$, forcing the second factor to be zero.Therefore $(\sin A-\sin B)^2+(\sin B-\sin C)^2+(\sin C-\sin A)^2 = 0$, which gives $\sin A = \sin B = \sin C$.Since angles are in $(0,\pi)$, this implies $A = B = C = 60^\circ$; the triangle is equilateral.Thus sides satisfy $a = b = c$.Now compute the determinant $D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$.Because $a=b=c$, all three rows are identical, so $D = 0$.(Alternatively, any column operation leads to a row of zeros.)Answer:Option D: $0$.