Let the set of all $$a \in \mathbb{R}$$ such that the equation $$\cos 2x + a \sin x = 2a - 7$$ has a solution be $$[p, q]$$ and $$r = \tan 9° - \tan 27° - \frac{1}{\cot 63°} + \tan 81°$$, then $$pqr$$ is equal to _______.

First, find $$[p, q]$$ for the equation $$\cos 2x + a\sin x = 2a - 7$$.

Using $$\cos 2x = 1 - 2\sin^2 x$$:

$$1 - 2\sin^2 x + a\sin x = 2a - 7$$

$$-2\sin^2 x + a\sin x + 8 - 2a = 0$$

$$2\sin^2 x - a\sin x + 2a - 8 = 0$$

Let $$t = \sin x \in [-1, 1]$$:

$$2t^2 - at + 2a - 8 = 0$$

$$a(2 - t) = 8 - 2t^2 = 2(4 - t^2) = 2(2-t)(2+t)$$

$$a = \frac{2(2-t)(2+t)}{2-t} = 2(2+t) = 4 + 2t$$ (for $$t \neq 2$$, which is always true since $$t \in [-1,1]$$).

For $$t \in [-1, 1]$$: $$a = 4 + 2t \in [2, 6]$$.

So $$p = 2$$, $$q = 6$$.

Now find $$r = \tan 9° - \tan 27° - \frac{1}{\cot 63°} + \tan 81°$$.

$$\frac{1}{\cot 63°} = \tan 63°$$.

$$r = \tan 9° - \tan 27° - \tan 63° + \tan 81°$$

$$= (\tan 9° + \tan 81°) - (\tan 27° + \tan 63°)$$

$$= (\tan 9° + \cot 9°) - (\tan 27° + \cot 27°)$$

$$= \frac{1}{\sin 9° \cos 9°} - \frac{1}{\sin 27° \cos 27°}$$

$$= \frac{2}{\sin 18°} - \frac{2}{\sin 54°}$$

$$= \frac{2}{\frac{\sqrt{5}-1}{4}} - \frac{2}{\frac{\sqrt{5}+1}{4}}$$

$$= \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1}$$

$$= 8 \cdot \frac{(\sqrt{5}+1) - (\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = 8 \cdot \frac{2}{4} = 4$$

$$pqr = 2 \times 6 \times 4 = 48$$

The answer is $$\boxed{48}$$.