If the domain of the function $$f(x) = \sin^{-1}\left(\frac{x-1}{2x+3}\right)$$ is $$\mathbb{R} - (\alpha, \beta)$$, then $$12\alpha\beta$$ is equal to :

For $$f(x) = \sin^{-1}\left(\frac{x-1}{2x+3}\right)$$ to be defined, we need $$-1 \leq \frac{x-1}{2x+3} \leq 1$$ and $$2x+3 \neq 0$$.

Condition 1: $$\frac{x-1}{2x+3} \leq 1$$

$$\frac{x-1}{2x+3} - 1 \leq 0 \implies \frac{x-1-2x-3}{2x+3} \leq 0 \implies \frac{-x-4}{2x+3} \leq 0 \implies \frac{x+4}{2x+3} \geq 0$$

This holds when $$x \leq -4$$ or $$x > -\frac{3}{2}$$.

Condition 2: $$\frac{x-1}{2x+3} \geq -1$$

$$\frac{x-1}{2x+3} + 1 \geq 0 \implies \frac{x-1+2x+3}{2x+3} \geq 0 \implies \frac{3x+2}{2x+3} \geq 0$$

This holds when $$x \leq -\frac{3}{2}$$ or $$x \geq -\frac{2}{3}$$.

Intersection of both conditions:

$$\left((-\infty, -4] \cup (-\frac{3}{2}, \infty)\right) \cap \left((-\infty, -\frac{3}{2}) \cup [-\frac{2}{3}, \infty)\right)$$

$$= (-\infty, -4] \cup [-\frac{2}{3}, \infty)$$

So the domain is $$(-\infty, -4] \cup [-\frac{2}{3}, \infty) = \mathbb{R} - (-4, -\frac{2}{3})$$.

Thus $$\alpha = -4, \beta = -\frac{2}{3}$$.

$$12\alpha\beta = 12 \times (-4) \times (-\frac{2}{3}) = 12 \times \frac{8}{3} = 32$$.

The correct answer is Option 1: 32.