MCQ
The points represented  by $\vec a,\vec b,\vec c,\vec d$ are coplanar and $\left( {\sin A} \right)\vec a + \left( {2\sin 2B} \right)\vec b + \left( {3\sin 3C} \right)\vec c - 4\vec d = \vec 0$ then the least value of $\frac{{21}}{8}\left( {{{\sin }^2}A + {{\sin }^2}2B + {{\sin }^2}3C} \right)$ is
  • A
    $1$
  • B
    $2$
  • C
    $4$
  • $3$

Answer

Correct option: D.
$3$
d
clearly $sinA +2sin2B+3sin3C=4$

$⇒$ ${\sin ^2}A + {\sin ^2}2B + {\sin ^2}3C \geqslant \frac{{{4^2}}}{{14}}$

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