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 perpendicular drawn from the points on L to the plane $P_1$. Which of the following points lie(s) on M ?

(a) $\left(0, -\frac{5}{6}, -\frac{2}{3}\right)$

(b) $\left( -\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right)$

(c) $\left(0, -\frac{5}{6}, -\frac{2}{3}\right)$

(d) $\left( -\frac{1}{3}, 0, \frac{2}{3}\right)$

(a), (b) 

The line L passes through the origin and is parallel to the planes $P_1$ and $P_2$.

Let the equation of line L be $\frac{x-0}{a}=\frac{y-0}{b}=\frac{z-0}{c}$

It is parallel to the planes $P_1$ and $P_2$.

$∴ a + 2b - c = 0 $ and $ 2a - b + c = 0 $

$⇒ \frac{a}{1}= \frac{b}{-3}= \frac{c}{-5}⇒ \frac{a}{-1}= \frac{b}{3}= \frac{c}{5}$

So, the equation of line L is

$\frac{x}{-1}=\frac{y}{3}=\frac{z}{5}$ ........(i)

The coordinates of any points on line (i) are (-γ, 3γ, 5γ ). The coordinates of the foot of perpendicular from (-γ, 3γ, 5γ) on the plane $P_1$ are given by

$\frac{x+γ}{1}=\frac{y-3γ}{2}=\frac{3-5γ}{-1}=-\frac{(-γ+6γ-5γ+1)}{1+4+1}$

or, $\frac{x+γ}{1}=\frac{y-3γ}{2}=\frac{3-5γ}{-1}=-\frac{1}{6}$

$⇒ x= -γ -\frac{1}{6}, y = 3 γ -\frac{2}{6}, z= 5γ +\frac{1}{6}$

So, the coordinates of the foot of perpendicular from (-γ, 3γ, 5γ) are $\left(-γ -\frac{1}{6}, 3γ-\frac{2}{6}, 5γ+\frac{1}{6}\right)$.

The clearly, M is the locus of points $\left(-γ -\frac{1}{6}, 3γ-\frac{2}{6}, 5γ+\frac{1}{6}\right)$, where γ is a variable. We observe that the coordinates in options (a) and (b) satisfy for $γ = -\frac{1}{6}$ and γ = 0 respectively.