$\text{ If for a }>0,\text{ the feet of perpendicular from the points }\mathrm{A}(\mathrm{a},-2\mathrm{a},3)\text{ and }B(0,4,5)\text{ on the plane }lx$+ my + nz = 0 are points C(0, -a, -1) and D respectively, then the length of line segment CD is equal to :

$\text{ Direction ratios of the line }\mathrm{AC}\text{ are }⟨-a,a,-4⟩$

Hence the equation of the plane which is passing through the origin and having normal 

$\text{ direction ratios }⟨-a,a,-4⟩\text{ is }ax-ay+4z=0$

$\text{ The point }C(0,-a,-1)\text{ lies on the above plane, hence }{a}^2-4=0\Rightarrow a=2$

$\text{ Equation of the plane is }2x-2y+4z=0\Rightarrow x-y+2z=0$

$\text{ The projection of }B(0,4,5)\text{ on the above plane is }$

$\frac{h}{1}=\frac{k-4}{-1}=\frac{l-5}{2}=-\frac{6}{6}\Rightarrow h=-1,k=5,l=3$

$\begin{array}{l}\text{ Hence, }D(-1,5,3)\text{ and }C(0,-2,-1)\\ \text{ Therefore, the distance between }C\text{ and }D\text{ is }\sqrt{1+49+16}=\sqrt{66}\end{array}$