Let $$|\cos\theta \cos(60° - \theta) \cos(60° + \theta)| \leq \frac{1}{8}$$, $$\theta \in [0, 2\pi]$$. Then, the sum of all $$\theta \in [0, 2\pi]$$, where $$\cos 3\theta$$ attains its maximum value, is :

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

So $$|\cos\theta \cos(60°-\theta)\cos(60°+\theta)| = \frac{1}{4}|\cos 3\theta| \leq \frac{1}{8}$$.

This gives $$|\cos 3\theta| \leq \frac{1}{2}$$, which is always true when $$|\cos 3\theta| \leq \frac{1}{2}$$.

But the question asks: the sum of all $$\theta \in [0, 2\pi]$$ where $$\cos 3\theta$$ attains its maximum value.

The maximum value of $$\cos 3\theta$$ is 1, which occurs when $$3\theta = 2n\pi$$, i.e., $$\theta = 0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi$$.

But we need $$|\cos 3\theta| \leq \frac{1}{2}$$ to be satisfied. At $$\cos 3\theta = 1$$, $$|\cos 3\theta| = 1 > \frac{1}{2}$$. So these points don't satisfy the constraint.

Given the constraint $$|\cos 3\theta| \leq \frac{1}{2}$$, the maximum of $$\cos 3\theta$$ subject to this constraint is $$\frac{1}{2}$$.

$$\cos 3\theta = \frac{1}{2}$$ when $$3\theta = \frac{\pi}{3} + 2n\pi$$ or $$3\theta = -\frac{\pi}{3} + 2n\pi = \frac{5\pi}{3} + 2n\pi$$.

For $$\theta \in [0, 2\pi]$$, $$3\theta \in [0, 6\pi]$$:

$$3\theta = \frac{\pi}{3}$$: $$\theta = \frac{\pi}{9}$$

$$3\theta = \frac{5\pi}{3}$$: $$\theta = \frac{5\pi}{9}$$

$$3\theta = \frac{7\pi}{3}$$: $$\theta = \frac{7\pi}{9}$$

$$3\theta = \frac{11\pi}{3}$$: $$\theta = \frac{11\pi}{9}$$

$$3\theta = \frac{13\pi}{3}$$: $$\theta = \frac{13\pi}{9}$$

$$3\theta = \frac{17\pi}{3}$$: $$\theta = \frac{17\pi}{9}$$

Sum = $$\frac{\pi}{9}(1+5+7+11+13+17) = \frac{54\pi}{9} = 6\pi$$.

The correct answer is Option 3: $$6\pi$$.