a, b, c are the lengths of three sides of a triangle ABC. If a, b, c are related by the relation a² + b² + c² = ab + bc + ca, then the value of (sin²A + sin²B + sin²C) is?

Solution (By Examveda Team)

$$\eqalign{ & {a^2} + {b^2} + {c^2} = ab + bc + ca \cr & \Rightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 \cr & \Rightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0 \cr & \Rightarrow {a^2} + {b^2} - 2ab + {b^2} + {c^2} - 2bc + {c^2} + {a^2} - 2ca = 0 \cr & \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0 \cr & \therefore a = b = c \cr & \vartriangle {\text{ABC}} = {\text{equilateral }}\vartriangle \cr & \therefore \angle {\text{A}} = \angle {\text{B}} = \angle {\text{C}} = {60^ \circ } \cr & {\text{So, }}{\sin ^2}{\text{A}} + {\sin ^2}{\text{B}} + {\sin ^2}{\text{C}} \cr & \Rightarrow {\sin ^2}{60^ \circ } + {\sin ^2}{60^ \circ } + {\sin ^2}{60^ \circ } \cr & \Rightarrow 3{\sin ^2}{60^ \circ } \cr & \Rightarrow 3 \times {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \cr & \Rightarrow \frac{9}{4} \cr} $$