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?

We need to find when $$\cos 2A + \cos 2B + \cos 2C$$ is least for a triangle.

Using the identity: $$\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A \cos B \cos C$$

This is minimized when $$\cos A \cos B \cos C$$ is maximized, which occurs when $$A = B = C = 60°$$ (equilateral triangle). The minimum value is $$-1 - 4 \times \frac{1}{8} = -\frac{3}{2}$$.

For an equilateral triangle with inradius $$r = 3$$:

The relationship between inradius and side is: $$r = \frac{a}{2\sqrt{3}}$$ $$\Rightarrow a = 2\sqrt{3} \times 3 = 6\sqrt{3}$$

Checking each option:

Option 1: Perimeter = $$3 \times 6\sqrt{3} = 18\sqrt{3}$$. This is correct.

Option 2: For an equilateral triangle: $$\sin 2A + \sin 2B + \sin 2C = 3\sin 120° = \frac{3\sqrt{3}}{2}$$. Also $$\sin A + \sin B + \sin C = 3\sin 60° = \frac{3\sqrt{3}}{2}$$. These are equal. Correct.

Option 3: The distance from the incenter M to each vertex is $$MA = \frac{r}{\sin(A/2)} = \frac{3}{\sin 30°} = 6$$. The angle $$\angle AMB = 90° + \frac{C}{2} = 90° + 30° = 120°$$. Therefore: $$\vec{MA} \cdot \vec{MB} = |MA||MB|\cos(\angle AMB) = 6 \times 6 \times \cos 120° = 36 \times (-\frac{1}{2}) = -18$$. Correct.

Option 4: Area = $$\frac{\sqrt{3}}{4} \times (6\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 108 = 27\sqrt{3}$$. But the option states $$\frac{27\sqrt{3}}{2}$$, which is NOT correct.

Therefore, the statement that is NOT correct is option 4: area of $$\triangle ABC = \frac{27\sqrt{3}}{2}$$.