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)
A. $ \sqrt 2 $
B. 1
C. -1
D. $ - \sqrt 2 $

Hint: This is a three dimensional coordinate geometry of line question. To solve this question first we will derive the equation of the two lines from given data and their respective direction ratios. Then we will derive the direction ratios of line PQ and PR. Then as per given data and using perpendicular formula, we will get the value of $ \lambda $

Complete step-by-step answer:
Line $ {L_1} $ is given by, $ y = x,z = 1 $
Hence the equation of line $ {L_1} $ is $ \dfrac{x}{1} = \dfrac{y}{1} = \dfrac{{z - 1}}{0} = q $ (let)
 $ (1,1,0) $ is its direction ratios
 $ \Rightarrow x = q,y = q,z = 1 $
Let the coordinates of point Q be $ Q(q,q,1) $
Line $ {L_2} $ is given by $ y = - x,z = - 1 $
Hence the equation of line $ {L_2} $ is $ \dfrac{x}{1} = \dfrac{y}{{ - 1}} = \dfrac{{z + 1}}{0} = r $ (let)
 $ (1, - 1,0) $ is its direction ratios
 $ \Rightarrow x = r,y = - r,z = - 1 $
Let the coordinates of point R be $ R(r, - r, - 1) $
We know that the direction ratios of a line joining two points $ ({x_{1,}}{y_{1,}}{z_1}) $ and $ ({x_{2,}}{y_{2,}}{z_2}) $ are $ {x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1} $ ………………… Formula 1
Hence, the direction ratios of line joining PQ are, $ \lambda - q,\lambda - q,\lambda - 1 $
Again we know that two lines having direction ratios $ ({a_{1,}}{b_{1,}}{c_1}) $ and $ ({a_{2,}}{b_{2,}}{c_2}) $ are perpendicular if $ {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0 $ …………………… Formula 2
Now according to the question, $ PQ \bot {L_1} $
 $ \therefore $ $ 1(\lambda - q) + 1(\lambda - q) + 0(\lambda - 1) = 0 $
Simplifying the above equation we get,
 $ \lambda - q + \lambda - q = 0 $
 $ \Rightarrow 2\lambda = 2q $
 $ \Rightarrow \lambda = q $
The direction ratios of line joining PQ are, $ \lambda - \lambda ,\lambda - \lambda ,\lambda - 1 $
Similarly, the direction ratios of line joining PR are, $ \lambda - r,\lambda + r,\lambda + 1 $
Now according to the question, $ PR \bot {L_2} $
 $ \therefore $ $ 1(\lambda - r) + ( - 1)(\lambda + r) + 0(\lambda + 1) = 0 $
Simplifying the above equation we get,
 $ \lambda - r - \lambda - r = 0 $
 $ \Rightarrow r = o $
Hence, coordinates of R is $ (0,0, - 1) $
The direction ratios of line joining PR are, $ \lambda ,\lambda ,\lambda + 1 $
As $ \angle QPR = 90^\circ $ i.e. $ PQ \bot PR $
Then putting their direction ratios in formula 2 we get,
 $ (\lambda - \lambda )(\lambda - 0) + (\lambda - \lambda )(\lambda - 0) + (\lambda - 1)(\lambda + 1) = 0 $
Simplifying the above equation we get,
 $ (\lambda - 1)(\lambda + 1) = 0 $
 $ \lambda = 1 $ is rejected as P and Q are different points.
 $ \therefore $ $ \lambda = - 1 $

So, the correct answer is “Option C”.

Note: In 3-D coordinate geometry, the equation of line with direction vector l, m, n that passes through the point x’, y’, z’ is given by the formula, $ \dfrac{{x - x'}}{l} = \dfrac{{y - y'}}{m} = \dfrac{{z - z'}}{n} $
The direction ratios of a line joining two points $ ({x_{1,}}{y_{1,}}{z_1}) $ and $ ({x_{2,}}{y_{2,}}{z_2}) $ are $ {x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1} $ .
Two lines having direction ratios $ ({a_{1,}}{b_{1,}}{c_1}) $ and $ ({a_{2,}}{b_{2,}}{c_2}) $ are perpendicular if $ {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0 $ .
You should remember all the rules, properties and formulas coordinate geometry.
You might mistakenly take both the values, but according to the question, P and Q are two different points.