In a triangle, ABC if Δ=1 a b1 c a1 b c=0 then sin2A+sin2B+sin2C is
332
54
94
2
Evaluating along C1, we get
Δ=c2−ab+b2−ac+a2−bc=0⇒ (a−b)2+(b−c)2+(c−a)2=0⇒ a=b=c⇒A=B=C=π/3∴ sin2A+sin2B+sin2C=332