From a point $P(\lambda, \lambda, \lambda)$, perpendiculars $PQ$ and $PR$ are drawn respectively on the lines $y = x, z = 1$ and $y = -x, z = -1$. If $P$ is such that $\angle QPR$ is a right angle, then the possible value(s) of $\lambda$ is/(are)

$P(\lambda,\lambda,\lambda)$

$L_1:\ y=x,\ z=1,\qquad L_2:\ y=-x,\ z=-1$

PQ⊥ L₁, PR⊥ L₂, ∠ QPR=90^∘ To find: all λ with ∠ QPR=90^∘

$P \text{ lies on } x=y=z:\ \text{point } (\lambda,\lambda,\lambda)$

$L_1:\ \text{through } (0,0,1),\ \vec d_1=(1,1,0)$

$L_2:\ \text{through } (0,0,-1),\ \vec d_2=(1,-1,0)$

$1.\ \vec{PQ}\cdot\vec d_1=0 \Rightarrow Q \text{ on } L_1$

$2.\ \vec{PR}\cdot\vec d_2=0 \Rightarrow R \text{ on } L_2$

$3.\ \vec{PQ}\cdot\vec{PR}=0 \Rightarrow \lambda$

$Q=(t,t,1)\in L_1,\quad \vec{PQ}=(t-\lambda,\ t-\lambda,\ 1-\lambda)$

$\vec{PQ}\cdot\vec d_1=0 \Rightarrow 2(t-\lambda)=0 \Rightarrow t=\lambda$

$Q=(\lambda,\lambda,1)$

$R=(s,-s,-1)\in L_2,\quad \vec{PR}=(s-\lambda,\ -s-\lambda,\ -1-\lambda)$

$\vec{PR}\cdot\vec d_2=0 \Rightarrow 2s=0 \Rightarrow s=0$

$R=(0,0,-1)\quad (\text{independent of } \lambda)$

$\vec{PQ}=(0,\ 0,\ 1-\lambda)$

$\vec{PR}=(-\lambda,\ -\lambda,\ -1-\lambda)$

$\vec{PQ}\cdot\vec{PR}=(1-\lambda)(-1-\lambda)=\lambda^{2}-1$

$\lambda^{2}-1=0 \Rightarrow \lambda=\pm 1$

$\lambda=1:\ P(1,1,1)\in L_1 \Rightarrow Q=P,\ \text{angle undefined — rejected}$

$\lambda=-1 \Rightarrow \text{option } (3)$