Let the range of the function $$f(x)=6+16\cos x.\cos (\frac{\pi}{3}-x).\cos (\frac{\pi}{3}+x).\sin 3x.\cos 6x, x \in R\text{ be }[\alpha,\beta]$$.Then the distance of the point $$(\alpha,\beta)$$ from the line 3x + 4y + 12 = 0 is :

We need to find the range of $$f(x) = 6 + 16\cos x \cos(\frac{\pi}{3} - x)\cos(\frac{\pi}{3} + x)\sin 3x \cos 6x$$.

Simplify $$\cos x \cos(\frac{\pi}{3}-x)\cos(\frac{\pi}{3}+x)$$

Using the identity $$\cos A \cos(60°-A)\cos(60°+A) = \frac{1}{4}\cos 3A$$:

$$\cos x \cos(\frac{\pi}{3}-x)\cos(\frac{\pi}{3}+x) = \frac{1}{4}\cos 3x$$

Substitute

$$f(x) = 6 + 16 \cdot \frac{1}{4}\cos 3x \cdot \sin 3x \cdot \cos 6x$$

$$= 6 + 4\cos 3x \sin 3x \cos 6x$$

$$= 6 + 2\sin 6x \cos 6x$$ (using $$2\sin A\cos A = \sin 2A$$)

$$= 6 + \sin 12x$$

Find the range

Since $$-1 \leq \sin 12x \leq 1$$:

$$f(x) \in [6-1, 6+1] = [5, 7]$$

So $$\alpha = 5, \beta = 7$$.

Find distance from (5,7) to line 3x + 4y + 12 = 0

$$d = \frac{|3(5) + 4(7) + 12|}{\sqrt{9+16}} = \frac{|15+28+12|}{5} = \frac{55}{5} = 11$$

The correct answer is Option 1: 11.