From a point $P(\lambda, \lambda, \lambda)$, perpendicular $PQ$ 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)

$\text{Given: } P(\lambda,\lambda,\lambda)$

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

$PQ \perp L_1,\quad PR \perp L_2$

∠ QPR = 90^∘

$\text{To find: possible value(s) of } \lambda$

$L_1: (0,0,1),\ \vec{d}_1=(1,1,0)$

$L_2: (0,0,-1),\ \vec{d}_2=(1,-1,0)$

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

$\text{Plan: (i) parametrize } L_1,\ L_2$

$\text{(ii) } PQ\perp L_1,\ PR\perp L_2 \Rightarrow Q,\ R$

$\text{(iii) } (Q-P)\cdot(R-P)=0 \Rightarrow \lambda$

$L_1: (t,\ t,\ 1) \qquad L_2: (s,\ -s,\ -1)$

$Q=(t,t,1);\quad (P-Q)\cdot(1,1,0)=0$

$2(\lambda-t)=0 \Rightarrow t=\lambda \Rightarrow Q=(\lambda,\lambda,1)$

$R=(s,-s,-1);\quad (P-R)\cdot(1,-1,0)=0$

$-2s=0 \Rightarrow s=0 \Rightarrow R=(0,0,-1)$

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

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

∠ QPR=90^∘ ⇒ PQ⃗·PR⃗=0

$(0)(-\lambda)+(0)(-\lambda)+(1-\lambda)(-1-\lambda)=0$

$(1-\lambda)(-1-\lambda)=0 \Rightarrow \lambda^2-1=0$

$\lambda=\pm 1$

$\lambda=-1:\ P(-1,-1,-1),\ Q(-1,-1,1),\ R(0,0,-1)$

$\lambda=1:\ P(1,1,1)\ \text{lies on } L_1$

$\lambda = 1 \ \text{and}\ \lambda = -1$

$\Rightarrow \text{Options } (2)\ \text{and } (3)$