JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Three Dimensional Geometry

The plane passes through the origin, and $C$ is the foot of the perpendicular from $A$, so

is along the normal $(l,m,n)$. Taking $(l,m,n)=(-a,a,-4)$ and putting $C(0,-a,-1)$ in the plane gives $-a\cdot a+4=0$, so $a=2$ for $a>0$.
The normal is $(-2,2,-4)$, that is $(1,-1,2)$, and the plane is $x-y+2z=0$, with $A=(2,-4,3)$ and $C=(0,-2,-1)$.
The foot from $B(0,4,5)$ is

so $\overrightarrow{CD}=(-1,7,4)$ and

The length of the segment $CD$ is $\sqrt{66}$ units.