For $$\alpha, \beta \in \left(0, \frac{\pi}{2}\right)$$, let $$3\sin(\alpha + \beta) = 2\sin(\alpha - \beta)$$ and a real number $$k$$ be such that $$\tan\alpha = k\tan\beta$$. Then the value of $$k$$ is equal to

Given: $$3\sin(\alpha+\beta) = 2\sin(\alpha-\beta)$$.

Expanding: $$3(\sin\alpha\cos\beta + \cos\alpha\sin\beta) = 2(\sin\alpha\cos\beta - \cos\alpha\sin\beta)$$.

$$3\sin\alpha\cos\beta + 3\cos\alpha\sin\beta = 2\sin\alpha\cos\beta - 2\cos\alpha\sin\beta$$

$$\sin\alpha\cos\beta = -5\cos\alpha\sin\beta$$

$$\frac{\sin\alpha}{\cos\alpha} = \frac{-5\sin\beta}{\cos\beta}$$

$$\tan\alpha = -5\tan\beta$$

So $$k = -5$$.

The answer is Option (1): $$\boxed{-5}$$.