For a triangle ABC , the value of 2A + 2B + 2C is least. If its...

We know that,

$\cos 2A + \cos 2B + \cos 2C \ge - \dfrac{3}{2}$

i.e., $min\left(\cos 2A + \cos 2B + \cos 2C\right) = - \dfrac{3}{2}$

So,

$\cos 2A = \cos 2B = \cos 2C = - \dfrac{1}{2}$

$⇒ A = B = C = 60 °$

So, triangle $ABC$ is an equilateral triangle.

Let the side of triangle be $a$.

Also,

$r = \dfrac{∆}{s} = 3$

$⇒ \dfrac{2 ∆}{2s} = 3$

$⇒ \dfrac{2\left(\dfrac{\sqrt{3}}{4}{a}^{2}\right)}{3a} = 3$

$⇒ a = \dfrac{18}{\sqrt{3}} = 6\sqrt{3}$

So, perimeter is $= 3a = 18\sqrt{3}$ units.

Area$= \dfrac{\sqrt{3}}{4}{a}^{2} = \dfrac{81}{\sqrt{3}} = 27\sqrt{3}$ sq. units