Let $$\theta$$ be the angle between the planes $$P_1 = \vec{r} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 9$$ and $$P_2 = \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 15$$. Let L be the line that meets $$P_2$$ at the point (4, -2, 5) and makes angle $$\theta$$ with the normal of $$P_2$$. If $$\alpha$$ is the angle between L and $$P_2$$ then $$\tan^2\theta \cot^2\alpha$$ is equal to ______.

Consider the planes $$P_1: \vec{r}\cdot(\hat{i}+\hat{j}+2\hat{k})=9$$, i.e.\ $$x+y+2z=9$$ with normal $$\vec{n_1}=(1,1,2)$$, and $$P_2: \vec{r}\cdot(2\hat{i}-\hat{j}+\hat{k})=15$$, i.e.\ $$2x-y+z=15$$ with normal $$\vec{n_2}=(2,-1,1)$$.

To find the angle $$\theta$$ between the planes, note that $$\cos\theta=\frac{|\vec{n_1}\cdot\vec{n_2}|}{|\vec{n_1}||\vec{n_2}|} =\frac{|2-1+2|}{\sqrt{6}\,\sqrt{6}} =\frac{3}{6}=\frac12,$$ hence $$\theta=60°$$ and $$\tan\theta=\sqrt{3}\,.$$

Next, suppose a line L meets $$P_2$$ at the point $$(4,-2,5)$$ and makes an angle $$\theta=60°$$ with $$\vec{n_2}$$. Then the angle $$\alpha$$ between L and the plane $$P_2$$ is $$\alpha=90°-\theta=30°\,.$$

It follows that $$\tan^2\theta\;\cot^2\alpha =\tan^2 60°\;\cot^2 30° =3\cdot3=9,$$ so the required value is $$\boxed{9}\,.$$