Let a curve $$y = y(x)$$ pass through the point $$(3, 3)$$ and the area of the region under this curve, above the $$x$$-axis and between the abscissae $$3$$ and $$x (> 3)$$ be $$\left(\dfrac{y}{x}\right)^3$$. If this curve also passes through the point $$(\alpha, 6\sqrt{10})$$ in the first quadrant, then $$\alpha$$ is equal to ______.

We need to find $$\alpha$$ such that the curve $$y = y(x)$$ passes through $$(3, 3)$$ and $$(\alpha, 6\sqrt{10})$$, where the area under the curve from 3 to $$x$$ equals $$\left(\dfrac{y}{x}\right)^3$$. Since the area condition gives $$\int_3^x y(t)\,dt = \left(\dfrac{y}{x}\right)^3$$, differentiating both sides with respect to $$x$$ yields $$y = 3\left(\dfrac{y}{x}\right)^2 \cdot \dfrac{y'x - y}{x^2}$$. This simplifies to $$y = \dfrac{3y^2(xy' - y)}{x^4}$$ and hence $$x^4 = 3y(xy' - y)$$.

Substituting $$y = vx$$ gives $$xy' - y = x(v'x + v) - vx = v'x^2$$, so $$x^4 = 3(vx)(v'x^2) = 3vv'x^3$$. From this, $$x = 3v\,\dfrac{dv}{dx}$$ and therefore $$3v\,dv = x\,dx$$.

Integrating both sides gives $$\dfrac{3v^2}{2} = \dfrac{x^2}{2} + C$$, which leads to $$3v^2 = x^2 + 2C$$.

At $$x = 3$$, we have $$v = y/x = 1$$. Substituting gives $$3(1)^2 = 9 + 2C \implies 2C = -6 \implies C = -3$$, so $$3v^2 = x^2 - 6 \implies 3\left(\dfrac{y}{x}\right)^2 = x^2 - 6$$ and hence $$3y^2 = x^4 - 6x^2$$.

Substituting $$y = 6\sqrt{10}$$ and $$x = \alpha$$ yields $$3(6\sqrt{10})^2 = \alpha^4 - 6\alpha^2$$, that is $$3 \times 360 = \alpha^4 - 6\alpha^2$$ and hence $$\alpha^4 - 6\alpha^2 - 1080 = 0$$.

Let $$u = \alpha^2$$: then $$u^2 - 6u - 1080 = 0$$ and $$u = \dfrac{6 \pm \sqrt{36 + 4320}}{2} = \dfrac{6 \pm \sqrt{4356}}{2} = \dfrac{6 \pm 66}{2}$$. Taking the positive root gives $$u = 36$$, so $$\alpha^2 = 36$$ and hence $$\alpha = 6$$ (first quadrant).

The correct answer is $$\boxed{6}$$.