UPSC CDS 2 2023 Math Paper

Concept:The given expression can be rewritten as a sum of squares, which is always non‑negative.For the sum to be zero, each square term must individually be zero.Explanation:We start with the equation:$$(a+b+c)^{2}-3(ab+bc+ca)+3(x^{2}+y^{2}+z^{2})=0$$Use the algebraic identity:$$(a+b+c)^{2}-3(ab+bc+ca)=\frac{1}{2}\big[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\big]$$This expression is always greater than or equal to zero.Also, $3(x^{2}+y^{2}+z^{2})\ge 0$ for real $x,y,z$.The entire left side is the sum of two non‑negative terms.Their sum is zero only when both terms are zero.First term zero implies $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$, so $a=b=c$.Second term zero implies $x^{2}+y^{2}+z^{2}=0$, so $x=y=z=0$.Thus the only possibility is $a=b=c$ and $x=y=z=0$.Answer:Option C: $a=b=c$, $x=y=z=0$