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 $M$?

In R³, 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 P₁ : x + 2y - z + 1 = 0 and P₂ : 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 P₁. Which of the following points lie(s) on M? $P_1: x + 2y - z + 1 = 0$ $P_2: 2x - y + z - 1 = 0$ $L \text{ passes through } O(0,0,0)$ $\text{all points of } L \text{ at constant distance from } P_1, P_2$ $M = \text{locus of feet of }\perp \text{ from } L \text{ to } P_1$ $\vec{n_1}=(1,2,-1),\qquad \vec{n_2}=(2,-1,1)$ $1.\ \text{constant distance} \Rightarrow \vec{d}\cdot\vec{n_1} = \vec{d}\cdot\vec{n_2} = 0$ $2.\ \vec{d} = \vec{n_1} \times \vec{n_2}$ $3.\ M = \text{projection of } L \text{ on } P_1;\ \text{check points}$ $\vec{r}=t\vec{d} \text{ on } L$ $\text{dist to } P_1 = \frac{|t(\vec{d}\cdot\vec{n_1})+1|}{\sqrt6} \Rightarrow \vec{d}\cdot\vec{n_1}=0$ $\text{and } \vec{d}\cdot\vec{n_2}=0$ $\vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & -1 & 1 \end{vmatrix}$ $\vec{d} = (1,-3,-5)$ $L \parallel P_1 \Rightarrow M = \text{proj}_{P_1}(L)$ $M \text{ is a line} \parallel L,\quad \vec{d_M}=(1,-3,-5)$ $Q_0 = O - \frac{1}{\|\vec{n_1}\|^2}\vec{n_1} = -\frac{1}{6}(1,2,-1)$ $Q_0 = \left(-\frac{1}{6}, -\frac{1}{3}, \frac{1}{6}\right)$ $M: \vec{r} = \left(-\frac{1}{6},-\frac{1}{3},\frac{1}{6}\right) + s(1,-3,-5)$ $(1):\ s=\frac{1}{6} \Rightarrow y=-\frac{1}{3}-\frac{3}{6}=-\frac{5}{6},\ z=\frac{1}{6}-\frac{5}{6}=-\frac{2}{3}$ $\Rightarrow \text{on } M\ (\text{TRUE})$ $(2):\ s=0 \Rightarrow Q_0 \in M$ $\Rightarrow \text{TRUE}$ $(3):\ s=-\frac{2}{3} \Rightarrow y=-\frac{1}{3}+2=\frac{5}{3} \neq 0$ $\Rightarrow \text{FALSE}$ $(4):\ s=-\frac{1}{6} \Rightarrow y=-\frac{1}{3}+\frac{1}{2}=\frac{1}{6} \neq 0$ $\Rightarrow \text{FALSE}$ $\text{True: (1) and (2)}$ $\boxed{\text{Answer: (1), (2)}}$