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 of  $\lambda$ is 

Let the line L1 be x = y, z = 1. Any point on this line is of the form (a, a, 1) 
  $(\lambda ,\lambda ,1)$
Let coordinates of Q be (a, a, 1)
Direction rations of PQ are
$\lambda –a,\lambda –a,\lambda –1$ .
As  $PQ\perp L_1$
$1(\lambda –a)+1(\lambda –a)+0(\lambda –1)=0\Rightarrow a=\lambda$ 
 .
 Coordinates of Q are  .
Direction rations of PQ are  .
Let line L2 be x = – y, z = – 1. Any point on L2 is of the form (b, – b, – 1).
Let coordinates of R be (b, – b, – 1) Direction rations of PR be  $\lambda –b,\lambda +b,\lambda +1$
As $PR\perp L_2$ , we get 
$0,0,\lambda –1$ 
 Thus coordinates of R (0, 0, – 1)
Direction ratios of PR are  $\lambda ,\lambda ,\lambda +1$
As  $QP\perp PR,(\lambda +1)(\lambda –1)=0$
$\Rightarrow \lambda =–1,1$ 
Reject$\lambda =1$  as for $\lambda =1$and Q coincide.