Let $$Y = Y(X)$$ be a curve lying in the first quadrant such that the area enclosed by the line $$Y - y = Y'(x)(X - x)$$ and the co-ordinate axes, where $$(x, y)$$ is any point on the curve, is always $$\frac{-y^2}{2Y'(x)} + 1$$, $$Y'(x) \neq 0$$. If $$Y(1) = 1$$, then $$12Y(2)$$ equals ________.

The tangent line at $$(x,y)$$ is $$Y - y = Y'(x)(X - x)$$. This meets the axes at:

X-axis ($$Y=0$$): $$X = x - y/Y'(x)$$. Y-axis ($$X=0$$): $$Y = y - xY'(x)$$.

Area of triangle with these intercepts and origin = $$\frac{1}{2}|x - y/Y'||y - xY'|$$.

Given this equals $$-\frac{y^2}{2Y'} + 1$$. Since the curve is in the first quadrant with the tangent having appropriate orientation:

$$\frac{1}{2}\left(x - \frac{y}{Y'}\right)(y - xY') = -\frac{y^2}{2Y'} + 1$$

$$\frac{1}{2}\left(xy - x^2Y' - \frac{y^2}{Y'} + xy\right) = -\frac{y^2}{2Y'} + 1$$

$$xy - \frac{x^2Y'}{2} - \frac{y^2}{2Y'} = -\frac{y^2}{2Y'} + 1$$

$$xy - \frac{x^2Y'}{2} = 1$$

$$Y' = \frac{2(xy-1)}{x^2} = \frac{2y}{x} - \frac{2}{x^2}$$

This is linear: $$Y' - \frac{2}{x}Y = -\frac{2}{x^2}$$. IF = $$e^{-2\ln x} = x^{-2}$$.

$$\frac{d}{dx}(Yx^{-2}) = -\frac{2}{x^4}$$. $$Yx^{-2} = \frac{2}{3x^3} + C$$.

$$Y = \frac{2}{3x} + Cx^2$$. $$Y(1) = 1$$: $$1 = 2/3 + C \Rightarrow C = 1/3$$.

$$Y = \frac{2}{3x} + \frac{x^2}{3}$$. $$Y(2) = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$$.

$$12Y(2) = 12 \times 5/3 = 20$$.

Therefore, the answer is $$\boxed{20}$$.