In $R^3$ , let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance 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?

As L is at a constant distance from P1 and P2, L is parallel to both P1 and P2. If direction ratios of L are 
a, b, c then
a + 2b – c = 0
2a – b + c = 0
 $$\begin{gathered} \Rightarrow \frac{a}{2–1}=\frac{–b}{1+2}=\frac{c}{–1–4} \\ \Rightarrow \frac{a}{1}=\frac{b}{–3}=\frac{c}{–5} \end{gathered}$$
So equation of line L is 
  $\frac{x}{1}=\frac{y}{–3}=\frac{z}{–5}=\lambda \left(say\right)$
A point on L is  $(\lambda ,–3\lambda ,–5\lambda )$
Let $I(\alpha ,\beta ,\gamma )$ be foot of perpendicular of  $P(\lambda ,–3\lambda ,–5\lambda )$ on the plane
 $P_1:x+2y–z+1=0$
Then IP is normal to the plane P1. Therefore 
  $$\begin{gathered} \frac{\alpha –\lambda }{1}=\frac{\beta +3\lambda }{2}=\frac{\gamma +5\lambda }{–1}=k\left(say\right) \\ \Rightarrow \alpha =\lambda +k,\beta =2k–3\lambda ,\gamma =–k–5\lambda \end{gathered}$$
 As I lies on P1
 $$\begin{gathered} (\lambda +k)+2(2k–3\lambda )–(–k–5\lambda )+1=0 \\ \Rightarrow 6k=–1\Rightarrow k=–\frac{1}{6} \end{gathered}$$
 Thus,  $\alpha =\lambda –\frac{1}{6},\beta =–3\lambda –\frac{1}{3},\gamma =–5\lambda +\frac{1}{6}$
 $\therefore M=\{(\lambda –\frac{1}{6},–3\lambda –\frac{1}{3},–5\lambda +\frac{1}{6})\lambda \in R\}$
For  $\lambda =0,(–\frac{1}{6},–\frac{1}{3},\frac{1}{6})\in M$
And For   $\lambda =\frac{1}{6},(0,\frac{–5}{6},\frac{–2}{3})\in M$