Three lines $L_1 : r = \lambda\hat{i}, \lambda \in R$, $L_2 : r = \hat{k} + \mu\hat{j}, \mu \in R$ and $L_3 : \vec{r} = \hat{i} + \hat{j} + \nu\hat{k}, \nu \in R$ are given. For which point(s) $Q$ on $L_2$ can we find a point $P$ on $L_1$ and a point $R$ on $L_3$ so that $P$, $Q$ and $R$ are collinear?

Three lines L₁ : r = λî, λ ∈ R, L₂ : r = k̂ + μĵ, μ ∈ R and L₃ : r⃗ = î + ĵ + νk̂, ν ∈ R are given. For which point(s) Q on L₂ can we find a point P on L₁ and a point R on L₃ so that P, Q and R are collinear? $L_1: \vec{r} = \lambda\hat{i}$ $L_2: \vec{r} = \hat{k} + \mu\hat{j}$ $L_3: \vec{r} = \hat{i} + \hat{j} + \nu\hat{k}$ $Q \in L_2,\quad P \in L_1,\quad R \in L_3$ $P,\ Q,\ R \text{ collinear}$ $\vec{d_1} = \hat{i},\quad \vec{d_2} = \hat{j},\quad \vec{d_3} = \hat{k}$ $L_1 \ni (0,0,0),\ L_2 \ni (0,0,1),\ L_3 \ni (1,1,0)$ $Q = (0,\ q,\ 1) \in L_2$ $\text{Options: } q = -\frac{1}{2},\ 0,\ 1,\ \frac{1}{2}$ $1.\ \text{plane through } L_1 \text{ and } Q$ $2.\ R = \text{plane} \cap L_3$ $3.\ QR \text{ meets } L_1 \text{ (same plane, unless parallel)}$ $\vec{n} = \hat{i} \times (0,q,1) = (0,-1,q)$ $\text{plane}: -y + qz = 0 \Rightarrow y = qz$ $R = (1,1,\nu),\quad 1 = q\nu \Rightarrow \nu = \frac{1}{q}$ $R = \left(1,\ 1,\ \frac{1}{q}\right),\quad q \neq 0$ $\vec{QR} = \left(1,\ 1-q,\ \frac{1}{q}-1\right)$ $\vec{QR} \parallel \hat{i} \Leftrightarrow 1-q=0 \Rightarrow q=1$ $QR \text{ and } L_1 \text{ lie in the same plane}$ $\text{Condition: } q \neq 0 \ \text{and}\ q \neq 1$ $q=0:\ \text{plane } y=0$ $L_3 \text{ has } y=1 \Rightarrow \text{no } R \Rightarrow \text{FALSE}$ $q=1:\ R=(1,1,1),\ \vec{QR}=(1,0,0) \parallel L_1$ $\Rightarrow \text{FALSE}$ $q=-\frac{1}{2}:\ R=(1,1,-2)$ $P=\left(\frac{1}{3},0,0\right) \Rightarrow \text{TRUE}$ $q=\frac{1}{2}:\ R=(1,1,2)$ $P=(-1,0,0),\ Q = \text{midpoint of } PR \Rightarrow \text{TRUE}$ $\text{True: (1) and (4)}$ $\boxed{\text{Answer: (1), (4)}}$