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 $M$ ?
$(A)(0,\frac{-5}{6},\frac{-2}{3})\hspace{1cm}(B)(\frac{-1}{6},\frac{-1}{3},\frac{1}{6})\hspace{1cm}(C)(\frac{-5}{6},0,\frac{1}{6})\hspace{1cm}(D)(\frac{-1}{3},0,\frac{2}{3})$
My try:
Line $L$ will be parallel to the line of intersection of $P_1$ and $P_2$.
Let $a, b$ and $c$ be the direction ratios of line $L$
$\Rightarrow a+2b-c=0,2a-b+c=0$
$\Rightarrow a:b:c::1:-3:-5$
Equation of line $L$ is $\frac{x-0}{1}=\frac{y-0}{-3}=\frac{z-0}{-5}$
But now i am stuck.How to further solve it?This is multiple options correct type question.
Please help me.
As you wrote, $L$ can be written as $\frac{x-0}{1}=\frac{y-0}{-3}=\frac{z-0}{-5}$, so a point $Q$ on $L$ can be expressed as $Q(t,-3t,-5t)$.
Let $M(p,q,r)$. Then, since $M$ is on $P_1$, $$p+2q-r+1=0\tag1$$
Since $\vec{QM}=(p-t,q+3t,r+5t)$ is parallel to $(1,2,-1)$, we have $$p-t:q+3t:r+5t=1:2:-1$$ From this, we have $$q=2p-5t,\quad r=-p-4t\tag2$$
Then, from $(1)$, we have $p=t-\frac 16$, so it follows from this and $(2)$ that $$M\left(t-\frac 16,-3t-\frac 13,-5t+\frac 16\right).$$
Thus, $(A)(B)$ are the only correct options.