The plane $ax + by = 0$ is rotated about its line of intersection with the plane $z = 0$ through an angle $\alpha$. Prove that the equation of the plane in its new position is $ax + by \pm (\sqrt{a^{2} + b^{2}} \tan \alpha) z = 0$.

(A) The equation of the given plane is $ax + by = 0$ ... $(i)$.
The equation of the plane $z = 0$ is ... $(ii)$.
Any plane passing through the line of intersection of planes $(i)$ and $(ii)$ can be written as $ax + by + kz = 0$ ... $(iii)$.
The normal to plane $(i)$ is $\vec{n_1} = (a, b, 0)$ and the normal to plane $(iii)$ is $\vec{n_2} = (a, b, k)$.
The angle $\alpha$ between these two planes is given by $\cos \alpha = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.
$\cos \alpha = \frac{|a^2 + b^2 + 0|}{\sqrt{a^2 + b^2} \sqrt{a^2 + b^2 + k^2}} = \frac{a^2 + b^2}{\sqrt{a^2 + b^2} \sqrt{a^2 + b^2 + k^2}} = \sqrt{\frac{a^2 + b^2}{a^2 + b^2 + k^2}}$.
Squaring both sides: $\cos^2 \alpha = \frac{a^2 + b^2}{a^2 + b^2 + k^2}$.
$a^2 + b^2 + k^2 = \frac{a^2 + b^2}{\cos^2 \alpha} = (a^2 + b^2) \sec^2 \alpha$.
$k^2 = (a^2 + b^2) \sec^2 \alpha - (a^2 + b^2) = (a^2 + b^2) (\sec^2 \alpha - 1) = (a^2 + b^2) \tan^2 \alpha$.
Thus,$k = \pm \sqrt{a^2 + b^2} \tan \alpha$.
Substituting $k$ into equation $(iii)$,we get $ax + by \pm (\sqrt{a^2 + b^2} \tan \alpha) z = 0$.