In mathbb{R}^3,let L be a straight line passi | vedclass

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(A) The line $L$ passes through the origin $(0, 0, 0)$ and is equidistant from planes $P_1$ and $P_2$. Thus,$L$ must be parallel to the line of inters

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$(A, B)$

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(A) The line $L$ passes through the origin $(0, 0, 0)$ and is equidistant from planes $P_1$ and $P_2$. Thus,$L$ must be parallel to the line of intersection of $P_1$ and $P_2$.Let the direction ratios of $L$ be $(a, b, c)$. Since $L$ is parallel to the intersection of $P_1$ and $P_2$,its direction vector is parallel to $\vec{n}_1 	imes \vec{n}_2$,where $\vec{n}_1 = (1, 2, -1)$ and $\vec{n}_2 = (2, -1, 1)$.$\vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -1 \ 2 & -1 & 1 \end{vmatrix} = \hat{i}(2-1) - \hat{j}(1+2) + \hat{k}(-1-4) = (1, -3, -5)$.So,the equation of line $L$ is $\frac{x}{1} = \frac{y}{-3} = \frac{z}{-5} = k$.$M$ is the locus of the feet of the perpendiculars from points on $L$ to plane $P_1$. This is the projection of line $L$ onto plane $P_1$.The projection of a line onto a plane is a line. The point on $L$ at $k=0$ is $(0, 0, 0)$. The foot of the perpendicular from $(0, 0, 0)$ to $P_1: x+2y-z+1=0$ is given by $\frac{x-0}{1} = \frac{y-0}{2} = \frac{z-0}{-1} = -\frac{0+0-0+1}{1^2+2^2+(-1)^2} = -\frac{1}{6}$.Thus,the foot is $\left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}ight)$,which is point $(B)$.Since $M$ is a line passing through $(B)$ with direction vector equal to the projection of $(1, -3, -5)$ onto $P_1$,we check the points. Point $(A) \left(0, -\frac{5}{6}, -\frac{2}{3}ight)$ lies on $P_1$ because $0 + 2(-\frac{5}{6}) - (-\frac{2}{3}) + 1 = -\frac{5}{3} + \frac{2}{3} + 1 = 0$. Testing if the vector from $B$ to $A$ is parallel to the projection direction confirms $(A)$ lies on $M$.

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