Let $$f$$ be a differentiable function satisfying $$f(x) = \frac{2}{\sqrt{3}} \int_0^{\sqrt{3}} f\left(\frac{\lambda^2 x}{3}\right) d\lambda$$, $$x > 0$$ and $$f(1) = \sqrt{3}$$. If $$y = f(x)$$ passes through the point $$(\alpha, 6)$$, then $$\alpha$$ is equal to _______.

We are given $$f(x) = \frac{2}{\sqrt{3}} \int_0^{\sqrt{3}} f\left(\frac{\lambda^2 x}{3}\right) d\lambda$$ for $$x > 0$$ and the condition $$f(1) = \sqrt{3}$$. To determine the form of $$f$$, assume $$f(x) = x^n$$ and substitute into the integral equation:

$$x^n = \frac{2}{\sqrt{3}} \int_0^{\sqrt{3}} \left(\frac{\lambda^2 x}{3}\right)^n d\lambda = \frac{2}{\sqrt{3}} \cdot \frac{x^n}{3^n} \int_0^{\sqrt{3}} \lambda^{2n} d\lambda$$

Evaluating the integral yields

$$x^n = \frac{2x^n}{\sqrt{3} \cdot 3^n} \cdot \frac{\lambda^{2n+1}}{2n+1}\Big|_0^{\sqrt{3}} = \frac{2x^n}{\sqrt{3} \cdot 3^n} \cdot \frac{(\sqrt{3})^{2n+1}}{2n+1}$$

Dividing by $$x^n$$ gives

$$1 = \frac{2}{\sqrt{3} \cdot 3^n} \cdot \frac{3^{(2n+1)/2}}{2n+1} = \frac{2 \cdot 3^{(2n+1)/2}}{\sqrt{3} \cdot 3^n \cdot (2n+1)} = \frac{2 \cdot 3^{n+1/2}}{3^{1/2} \cdot 3^n \cdot (2n+1)} = \frac{2}{2n+1}$$

Hence $$2n + 1 = 2$$, so $$n = \frac{1}{2}$$ and $$f(x) = c\,x^{1/2} = c\sqrt{x}$$. Using $$f(1) = \sqrt{3}$$ gives $$c = \sqrt{3}$$, and thus $$f(x) = \sqrt{3x}$$.

Finally, to find $$\alpha$$ such that $$f(\alpha) = 6$$, solve

$$\sqrt{3\alpha} = 6$$ $$3\alpha = 36$$ $$\alpha = 12$$ The answer is $$\boxed{12}$$.