In $R^3$let L be a striaht line passing through the origin . Suppose that al the points on L are at a constant distacne from the two planes $P_1:x+2y-z+1=0$ and $P_2:2x-y+z-1=0$ Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane $P_1$ Which of the following points lie (s) on M

Let $\overrightarrow{v}$ be the vector along L.
Clearly, line L is parallel to given planes

$p_1:x+2y-z+1=0$

$\text{ and }{p}_2:2x-y+z-1=0$

$\therefore \overrightarrow{v}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 1& 2& 1\\ 2& -1& 1\end{array}\right|=\widehat{i}-3\widehat{j}-5\widehat{k}$

Since L is passing through origin, any point on line

$L\text{ is }A(\lambda ,-3\lambda ,-5\lambda )$

Foot of perpendicular from A to $P_1$ is

$\frac{h-\lambda }{1}=\frac{k+3\lambda }{2}=\frac{l+5\lambda }{-1}=\frac{(\lambda -6\lambda +5\lambda +1)}{1+4+1}=-\frac{1}{6}$

$\therefore h=\lambda -\frac{1}{6},k=-3\lambda -\frac{1}{3},l=-5\lambda +\frac{1}{6}$

So, foot or point on locus M is

$\left(\lambda -\frac{1}{6},-3\lambda -\frac{1}{3},-5\lambda +\frac{1}{6}\right)$

So point (a) and (b) lies on this locus.