From a point P(λ, λ, λ), perpendiculars PQ and PR are drawn...

Line 1 : $\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z - 1}{0} - r, Q \left(r, r, 1\right) <br >Line 2 : \dfrac{x}{1} = \dfrac{y}{- 1} = \dfrac{z + 1}{0} =$k, R \left(k, - k, - 1\right)$<br >$\overset{\to}{PQ} = \left(\lambda - r\right)\overset{^}{i} + \left(\lambda - r\right)\overset{^}{j} + \left(\lambda - 1\right)\overset{^}{k}$<br >And$\lambda - r + \lambda - r = 0$as$\overset{\to}{PQ}$is$⊥ to {L}_{1}$<br >$⇒ 2\lambda = 2r ⇒ \lambda = r$<br >$\overset{\to}{PR} = \left(\lambda - k\right)\overset{^}{i} + \left(\lambda + k\right)\overset{^}{j} + \left(\lambda + 1\right)\overset{^}{k}$<br >And$\lambda - k - \lambda - k = 0$as$\overset{\to}{PR}$is$⊥ to {L}_{2}$<br >$⇒ k = 0$<br >So$PQ ⊥ PR$<br > \left(\lambda - r\right)$\left(\lambda - k\right)$+ \left(\lambda - r\right)$\left(\lambda + k\right)$+ \left(\lambda - 1\right)$\left(\lambda + 1\right)$= 0$
$⇒ \lambda = 1, - 1$
For $\lambda = 1$ as point P and Q coincide
⇒ \lambda = - 1 .