The equation of the planes parallel to the plane $$x - 2y + 2z - 3 = 0$$ which are at unit distance from the point $$(1, 2, 3)$$ is $$ax + by + cz + d = 0$$. If $$(b-d) = K(c-a)$$, then the positive value of $$K$$ is ________.

The given plane is $$x - 2y + 2z - 3 = 0$$. Any plane parallel to it has the form $$x - 2y + 2z + \lambda = 0$$ for some constant $$\lambda$$.

The distance from the point $$(1, 2, 3)$$ to this plane is $$\frac{|1(1) + (-2)(2) + 2(3) + \lambda|}{\sqrt{1^2 + (-2)^2 + 2^2}} = \frac{|1 - 4 + 6 + \lambda|}{\sqrt{1+4+4}} = \frac{|3 + \lambda|}{3}$$.

Setting this equal to 1: $$\frac{|3 + \lambda|}{3} = 1$$, so $$|3 + \lambda| = 3$$.

This gives $$3 + \lambda = 3$$ or $$3 + \lambda = -3$$, i.e., $$\lambda = 0$$ or $$\lambda = -6$$.

The two planes are $$x - 2y + 2z = 0$$ (when $$\lambda = 0$$) and $$x - 2y + 2z - 6 = 0$$ (when $$\lambda = -6$$).

For the plane $$x - 2y + 2z = 0$$: $$a = 1, b = -2, c = 2, d = 0$$. Then $$(b - d) = -2 - 0 = -2$$ and $$(c - a) = 2 - 1 = 1$$. So $$K = \frac{b-d}{c-a} = \frac{-2}{1} = -2$$.

For the plane $$x - 2y + 2z - 6 = 0$$: $$a = 1, b = -2, c = 2, d = -6$$. Then $$(b - d) = -2 - (-6) = 4$$ and $$(c - a) = 2 - 1 = 1$$. So $$K = \frac{b-d}{c-a} = \frac{4}{1} = 4$$.

The positive value of $$K$$ is $$4$$.