For a triangle $A B C$ , the value of $\cos 2 A + \cos 2 B + \cos 2 C$ is least. If its in radius is 3 and incentre is M, then which of the following is NOT correct?

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sin 2A + sin 2B + sin 2C = sin A + sin B + sin C

area of $∆$ ABC is $\frac{27 \sqrt{3}}{2}$

$\vec{MA} \cdot \vec{MB} =$ -18

Perimeter of $∆$ ABC is 18 $\sqrt{3}$

answer is A.

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Detailed Solution

Let K = cos 2A + cos 2B + cos 2C, is minimum if A = B = C = 60^o i.e K = -3/2
$∆$ ABC is equilateral triangle .
In radius r = 3 $\Rightarrow side = 2 \sqrt{3} r = 6 \sqrt{3}$
$Area = \frac{\sqrt{3}}{4} (side)^{2} = 27 \sqrt{3}$
Perimeter = 3a = 18 $\sqrt{3}$
$\vec{MA} \cdot \vec{MB} =$ 6.6(1/2) = -18
sin 2A + sin 2B + sin 2C = sin A +sin B + sin C