If in a $\Delta \,ABC$, ${\sin ^3}A + {\sin ^3}B + {\sin ^3}C = 3\,\,\sin A\,\,\sin \,\,B\,\,\sin \,\,C$, then $\left| {\,\begin{array}{*{20}{c}} a&b&c\\ b&c&a\\ c&a&b \end{array}\,} \right| = $

${\left( {a + b + c} \right)^3}$

$\left( {a + b + c} \right)\,\,\left( {ab + bc + ca} \right)$

JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

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Analyzing +4 | -1

If in a

\Delta \,ABC

{\sin ^3}A + {\sin ^3}B + {\sin ^3}C = 3\,\,\sin A\,\,\sin \,\,B\,\,\sin \,\,C

, then

\left| {\,\begin{array}{cccccccccccccccccccc} a&b&c\\ b&c&a\\ c&a&b \end{array}\,} \right| =

View answer and explanation

Correct option: A

\mathrm{0}

Option A (

\mathrm{0}

) is correct because the given condition implies that

A = B = C = 60^\circ

, making the determinant zero.