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In a triangle, $A B C$ if $Δ = |\begin{array}{lll} 1 a b \\ 1 c a \\ 1 b c \end{array}| = 0$ then $\sin^{2} A + \sin^{2} B + \sin^{2} C$ is

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Correct option is A

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∴ sin2⁡A+sin2⁡B+sin2⁡C=332

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In a triangle, ABC if Δ=1 a b1 c a1 b c=0 then sin2⁡A+sin2⁡B+sin2⁡C is

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In a triangle, $A B C$ if $Δ = |\begin{array}{lll} 1 a b \\ 1 c a \\ 1 b c \end{array}| = 0$ then $\sin^{2} A + \sin^{2} B + \sin^{2} C$ is