In ${\mathbf{R}}^3$, 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 $P_1$ and $P_2$, $L$ is parallel to both $P_1$ and $P_2$. 

If direction ratios of $L$ are $a, b, c$ then

$\begin{array}{l}a+2b-c=0\\ 2a-b+c=0\\ \Rightarrow \frac{a}{2-1}=\frac{-b}{1+2}=\frac{c}{-1-4}\\ \Rightarrow \frac{a}{1}=\frac{b}{-3}=\frac{c}{-5}\end{array}$

So equation of line $L$ is

$\frac{x}{1}=\frac{y}{-3}=\frac{z}{-5}=\lambda$ (say)

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

then $IP$ is normal to the plane $P_1$. Therefore

$\frac{\alpha -\lambda }{1}=\frac{\beta +3\lambda }{2}=\frac{\gamma +5\lambda }{-1}=k$ (say)

$\Rightarrow \alpha =\lambda +k,\beta =2k-3\lambda ,\gamma =-k-5\lambda$

As $I$ lies on $P_1$

$\begin{array}{l}(\lambda +k)+2(2k-3\lambda )-(-k-5\lambda )+1=0\\ \Rightarrow 6k=-1\Rightarrow k=-\frac{1}{6}\end{array}$

Thus, $\alpha =\lambda -\frac{1}{6},\beta =-3\lambda -\frac{1}{3},\gamma =-5\lambda +\frac{1}{6}$

$\therefore M=\left\{\left(\lambda -\frac{1}{6},-3\lambda -\frac{1}{3},-5\lambda +\frac{1}{6}\right)\lambda \in \mathbf{R}\right\}$

For $\lambda =0,\left(-\frac{1}{6},-\frac{1}{3},\frac{1}{6}\right)\in M$

and for $\lambda =\frac{1}{6},\left(0,\frac{-5}{6},\frac{-2}{3}\right)\in M$