For $$\alpha, \beta, \gamma \neq 0$$. If $$\sin^{-1}\alpha + \sin^{-1}\beta + \sin^{-1}\gamma = \pi$$ and $$(\alpha + \beta + \gamma)(\alpha - \gamma + \beta) = 3\alpha\beta$$, then $$\gamma$$ equal to

We need to find $$\gamma$$ given that $$\sin^{-1}\alpha + \sin^{-1}\beta + \sin^{-1}\gamma = \pi$$ and $$(\alpha + \beta + \gamma)(\alpha - \gamma + \beta) = 3\alpha\beta$$.

Simplifying the second condition, we write

$$(\alpha + \beta + \gamma)(\alpha + \beta - \gamma) = 3\alpha\beta$$

This is a difference of squares pattern:

$$(\alpha + \beta)^2 - \gamma^2 = 3\alpha\beta$$

which expands to

$$\alpha^2 + 2\alpha\beta + \beta^2 - \gamma^2 = 3\alpha\beta$$

and can be rewritten as

$$\alpha^2 - \alpha\beta + \beta^2 = \gamma^2 \quad \ldots (*)$$

The cosine rule states: $$c^2 = a^2 + b^2 - 2ab\cos C$$. Comparing with (*) in the form $$\gamma^2 = \alpha^2 + \beta^2 - \alpha\beta$$, we get

$$2\alpha\beta \cos C = \alpha\beta \implies \cos C = \frac{1}{2} \implies C = \frac{\pi}{3}$$.

If $$\sin^{-1}\alpha + \sin^{-1}\beta + \sin^{-1}\gamma = \pi$$, let $$A = \sin^{-1}\alpha$$, $$B = \sin^{-1}\beta$$ and $$C = \sin^{-1}\gamma$$ so that $$A + B + C = \pi$$ and $$A, B, C$$ are the angles of a triangle. By the sine rule, the sides are proportional to the sines of the corresponding angles, namely $$\alpha, \beta, \gamma$$. Since the angle opposite the side of length $$\gamma$$ is $$C = \frac{\pi}{3}$$, it follows that $$\sin^{-1}\gamma = \frac{\pi}{3}$$.

Hence

$$\sin^{-1}\gamma = \frac{\pi}{3} \implies \gamma = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$$.

The correct answer is Option 1: $$\frac{\sqrt{3}}{2}$$.