If $a + b + c = 0$ and $a^{2} + b^{2} + c^{2} = k (a^{2} - b c)$ then $k = . . . . . . . .$

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Solution

The correct option is B $2$
$a + b + c = 0$
$a = - b - c$ ..........(1)
$(a + b + c)^{2} = 0$
$a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 0$
$K (a^{2} - b c) + 2 (a b + b c + c a) = 0$
$2 (a b + b c + c a) = - K (a (- b - c) - b c)$
$⟹ 2 (a b + b c + c a) = K (a b + b c + c a)$
$⟹ K = 2$

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∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} + c^{2} & a b & a c a b & c^{2} + a^{2} & b c a c & b c & a^{2} + b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = K a^{2} b^{2} c^{2}

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