If in a $Δ A B C$ , ${sin}^{3} A + {sin}^{3} B + {sin}^{3} C = 3 sin A sin B sin C$ , then the value of determinant $∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣$ is equal to

0

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Solution

The correct option is B $0$
$∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = (a + b + c) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣$ $[R_{1} \to R_{1} + R_{2} + R_{3}]$

$= (a + b + c) (b c + c a + a b - a^{2} - b^{2} - c^{2})$
$= - (a^{3} + b^{3} + c^{3} - 3 a b c)$ .......... $(i)$

By Sine rule

$\frac{a}{S i n A} = \frac{b}{S i n B} = \frac{c}{S i n C} = R$

$∴ a = R S i n A, b = R S i n B, c = S i n C$

$f r o m (i)$

$= - 8 R^{3} ({sin}^{3} A + {sin}^{3} B + {sin}^{3} C - 3 sin A sin B sin C)$
$= - 8 R^{3} (3 sin A sin B sin C - 3 sin A sin B sin C)$
$= 0$

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