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

-1

The equations of the given lines are

$L_1 : y = x, z = 1$     or, $\frac{x}{1}=\frac{y}{1}= \frac{z-1}{0}$ .......(i)

$L_2 :y = -x, z = -1 $ or $ \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}$ .....(ii)

Let the coordinates Q and R be given by

$\frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}= r_1 $  and   $ \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=r_2$ respectively

Thus, the coordinates of Q and R be $(r_1, r_1, 1)$ and $(r_2, -r_2, -1)$ respectively.

$∴ \vec{PQ}=(r_1-λ)\hat{i} +(r_1- λ) \hat{j} + (1-λ)\hat{k}$

and, $\vec{PR}= (r_2- λ)\hat{i} + (r_2- λ) \hat{j} + (-1 - λ) \hat{k}$

It is given that PQ is perpendicular to $L_1$ and PR is perpendicular to $L_2.$

$∴ \vec{PQ}.(\hat{i} + \hat{j} + 0\hat{k})= 0 $ and $ \vec{PR}. (\hat{i} - \hat{j} + 0\hat{k})= 0 $

$⇒ r_1 - λ + r_1 - λ + (1- λ) = 0 $

and $(r_2 - λ) - (r_2-λ)+0(-1-λ)= 0 $

$⇒r_1 = λ $ and $r_2 = 0 $

$∴ \vec{PQ}= (1- λ) \hat{k} $ and $\vec{PR}= -λ \hat{i} - λ\hat{j} + (-1- λ) \hat{k}$

Also, $ \vec{PQ}⊥ \vec{PR}= 0 $

$⇒ (1- λ) (-1-λ) = 0 ⇒ λ =  ±1$

For $λ = 1, \vec{PQ}= \vec{0}$. So, P and Q coincide.

$∴ λ = - 1$

Hence, there is only one value of λ equal to -1.