The foot of the perpendicular drawn from the point (4, 2, 3) to the line joining the points (1, -2, 3) and (1, 1, 0) lies on the plane: 

Line AB is 

$\frac{x-1}{0}=\frac{y+2}{3}=\frac{z-3}{-3}\Rightarrow \frac{x-1}{0}=\frac{y+2}{1}=\frac{z-3}{-1}=t$

$\text{ General point on the above line is }\mathrm{Q}=(1,\mathrm{t}-2,-\mathrm{t}+3)$

$\text{ Direction rations of }\overrightarrow{\mathrm{PQ}}=(-3,\mathrm{t}-4,-\mathrm{t})\text{ and suppose that }Q\text{ is foot of the perpendicular }$

$\text{ of }P\text{ on }AB$

Hence, 

$\begin{array}{c}-3\left(0\right)+\left(t-4\right)1+\left(-t\right)\left(-1\right)=0\\ 0+t-4+t=0\\ 2t-4=0\\ t=2\end{array}$

$\text{ Therefore, the foot of the perpendicular of }P\text{ on the line is }Q(1,0,1)$

Verify the correct options for the plane