Answer:
Option A
Explanation:
In $\triangle$ PQR is right angled , at $\theta$
$\therefore$ PR2=PQ2+RQ2
$\Rightarrow (6-6)^{2}+(-10-10)^{2}+(\lambda-10)^{2}+[(1-6)^{2}+(0-10)^{2}+(-5-10)^{2}]$
= $[(1-6)^{2}+(0+10)^{2}+(-5-\lambda)^{2}]$
$\Rightarrow $ $ 400+\lambda^{2}+100-20\lambda=(25+100+225)+(25+100+25+\lambda^{2}+10\lambda)$
$\Rightarrow $ $ \lambda^{2}+20\lambda+500=350+150+10\lambda+\lambda^{2}$
$\Rightarrow $ $-20\lambda=10\lambda\Rightarrow30\lambda=0$
$\Rightarrow $ $ \lambda=0$