Product of Alpha Values for Given Limit Condition

To find the product of all possible values of $\alpha$ for which the given limit equals $2$, we evaluate:

$\lim_{x \to 0} \left( \frac{1 - \cos (\alpha x) \cos((\alpha + 1)x) \cos((\alpha + 2)x)}{\sin^2((\alpha + 1)x)} \right) = 2$

Using the Taylor series expansions for $\cos \theta$ and $\sin \theta$ around $\theta = 0$: $\cos \theta = 1 - \frac{\theta^2}{2} + O(\theta^4)$ $\sin \theta = \theta + O(\theta^3)$

Substituting these expansions into the terms in the limit: $\cos(\alpha x) = 1 - \frac{\alpha^2 x^2}{2} + O(x^4)$ $\cos((\alpha + 1)x) = 1 - \frac{(\alpha + 1)^2 x^2}{2} + O(x^4)$ $\cos((\alpha + 2)x) = 1 - \frac{(\alpha + 2)^2 x^2}{2} + O(x^4)$

Multiplying these three series together gives: $\cos (\alpha x) \cos((\alpha + 1)x) \cos((\alpha + 2)x) = 1 - \frac{x^2}{2} \left[ \alpha^2 + (\alpha + 1)^2 + (\alpha + 2)^2 \right] + O(x^4)$

Thus, the numerator becomes: $1 - \cos (\alpha x) \cos((\alpha + 1)x) \cos((\alpha + 2)x) = \frac{x^2}{2} \left[ \alpha^2 + (\alpha + 1)^2 + (\alpha + 2)^2 \right] + O(x^4)$

For the denominator: $\sin^2((\alpha + 1)x) = (\alpha + 1)^2 x^2 + O(x^4)$

Substituting the numerator and denominator back into the limit expression: $\lim_{x \to 0} \frac{\frac{x^2}{2} \left[ \alpha^2 + (\alpha + 1)^2 + (\alpha + 2)^2 \right] + O(x^4)}{(\alpha + 1)^2 x^2 + O(x^4)} = \frac{\alpha^2 + (\alpha + 1)^2 + (\alpha + 2)^2}{2(\alpha + 1)^2}$

We are given that this limit equals $2$: $\frac{\alpha^2 + (\alpha + 1)^2 + (\alpha + 2)^2}{2(\alpha + 1)^2} = 2$

Assuming $\alpha + 1 \neq 0$: $\alpha^2 + (\alpha + 1)^2 + (\alpha + 2)^2 = 4(\alpha + 1)^2$ $\alpha^2 + (\alpha + 2)^2 = 3(\alpha + 1)^2$

Expanding both sides: $\alpha^2 + (\alpha^2 + 4\alpha + 4) = 3(\alpha^2 + 2\alpha + 1)$ $2\alpha^2 + 4\alpha + 4 = 3\alpha^2 + 6\alpha + 3$

Rearranging into a standard quadratic equation: $\alpha^2 + 2\alpha - 1 = 0$

For this quadratic equation $\alpha^2 + 2\alpha - 1 = 0$, the roots are $\alpha_1, \alpha_2 = -1 \pm \sqrt{2}$. Since $\alpha + 1 = \pm\sqrt{2} \neq 0$, both values of $\alpha$ are valid.

By Vieta's formulas, the product of all possible values of $\alpha$ is: $\text{Product of roots} = \frac{c}{a} = \frac{-1}{1} = -1$

Hence, the correct option is C.