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

For every triangle $$ABC$$ we have the identity

$$\cos 2A+\cos 2B+\cos 2C = -1-4\cos A\cos B\cos C$$ $$-(1)$$

Proof of $$-(1)$$: write $$\cos 2A=1-2\sin^2A$$, add the three similar terms and use $$\sin^2A+\sin^2B+\sin^2C = 2+2\cos A\cos B\cos C$$ (obtained from the sine rule and $$A+B+C=\pi$$). Substitution yields $$-(1)$$.

To minimise the left-hand side of $$-(1)$$ we must maximise $$\cos A\cos B\cos C$$, because of the negative sign in front of it. By AM-GM, for fixed $$A+B+C=\pi$$,

$$\cos A\cos B\cos C \le \left(\dfrac{\cos A+\cos B+\cos C}{3}\right)^{3}$$

with equality only when $$A=B=C=\dfrac{\pi}{3}$$. Thus the product (and hence $$-(1)$$) attains its required extremum for the equilateral triangle:

$$A=B=C=\dfrac{\pi}{3},\qquad \cos 2A+\cos 2B+\cos 2C = -1-\dfrac12 = -\dfrac32$$.

Therefore the given triangle $$ABC$$ is equilateral.

Let the common side length be $$a$$. For an equilateral triangle

$$r=\dfrac{\sqrt{3}}{6}\,a,\qquad R=\dfrac{a}{\sqrt{3}}$$ $$-(2)$$

where $$r$$, $$R$$ are the in-radius and circum-radius respectively. The question gives $$r=3$$, so from $$-(2)$$

$$a = \dfrac{6r}{\sqrt{3}} = \dfrac{18}{\sqrt{3}} = 6\sqrt{3}.$$

Checking the statements one by one

Option A: Perimeter $$=3a=3(6\sqrt{3})=18\sqrt{3}$$, exactly as stated. Option A is true.

Option B: In an equilateral triangle $$\sin 2A=\sin 2B=\sin 2C=\sin\!\left(\dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}$$ and $$\sin A=\sin B=\sin C=\dfrac{\sqrt{3}}{2}$$. Hence

$$\sin 2A+\sin 2B+\sin 2C =3\left(\dfrac{\sqrt{3}}{2}\right) =\sin A+\sin B+\sin C,$$

so Option B is true.

Option C: The incentre $$M$$ of an equilateral triangle coincides with its centroid and circumcentre. Therefore

$$|MA| = |MB| = R = 6$$ (from $$-(2)$$) and $$\angle AMB = 120^\circ$$. The dot product is

$$\vec{MA}\cdot\vec{MB}=|MA||MB|\cos 120^\circ =6\cdot 6\cdot\left(-\dfrac12\right)=-18,$$

agreeing with Option C. Hence Option C is true.

Option D: Area of an equilateral triangle is

$$\Delta = \dfrac{\sqrt{3}}{4}\,a^{2} = \dfrac{\sqrt{3}}{4}\,(6\sqrt{3})^{2} = \dfrac{\sqrt{3}}{4}\,(108) = 27\sqrt{3}.$$

Option D states $$\dfrac{27\sqrt{3}}{2}$$, which is only half of the correct area. Therefore Option D is NOT correct.

Hence, the statement that is NOT correct is Option D.