Let $$\alpha$$ be the constant term in the binomial expansion of $$\left(\sqrt{x} - \frac{6}{x^{3/2}}\right)^n$$, $$n \leq 15$$. If the sum of the coefficients of the remaining terms in the expansion is $$649$$ and the coefficient of $$x^{-n}$$ is $$\lambda\alpha$$, then $$\lambda$$ is equal to _____.

The binomial expansion is $$\left(\sqrt{x} - \frac{6}{x^{3/2}}\right)^n$$ with $$n \leq 15$$.

The general term is:

$$T_{r+1} = \binom{n}{r}(\sqrt{x})^{n-r}\left(-\frac{6}{x^{3/2}}\right)^r = \binom{n}{r}(-6)^r \cdot x^{\frac{n-4r}{2}}$$

Finding $$n$$:

For the constant term ($$\alpha$$): $$\frac{n - 4r}{2} = 0 \Rightarrow r = \frac{n}{4}$$. Since $$r$$ must be a non-negative integer and $$n \leq 15$$, we have $$n \in \{4, 8, 12\}$$.

The sum of all coefficients (set $$x = 1$$): $$(1 - 6)^n = (-5)^n$$.

Given: sum of remaining coefficients = $$(-5)^n - \alpha = 649$$.

For $$n = 4$$: $$\alpha = \binom{4}{1}(-6)^1 = -24$$, and $$(-5)^4 - (-24) = 625 + 24 = 649$$. This works.

For $$n = 8$$ or $$n = 12$$: $$(-5)^n$$ is far too large. So $$n = 4$$ and $$\alpha = -24$$.

Finding $$\lambda$$:

We need the coefficient of $$x^{-n} = x^{-4}$$:

$$\frac{4 - 4r}{2} = -4 \Rightarrow r = 3$$

Coefficient of $$x^{-4}$$: $$\binom{4}{3}(-6)^3 = 4 \times (-216) = -864$$

Since the coefficient of $$x^{-n}$$ equals $$\lambda\alpha$$:

$$-864 = \lambda \times (-24) \Rightarrow \lambda = \frac{-864}{-24} = 36$$

The answer is $$\boxed{36}$$.