Consider the line $L\equiv \frac{x-1}{2}=\frac{y+2}{3}=\frac{z-7}{6}.$ Point $P\left(2,-\mathrm{5,0}\right)$ and $Q$ are such that $PQ$ is perpendicular to the line $L$ and the midpoint of $PQ$ lies on line $L,$ then coordinates of $Q$ are

• Correct answer
  A

  $\left(-4,-5,2\right)$
• B

  $\left(-\mathrm{3,0},1\right)$
• C

  $\left(\mathrm{1,6},2\right)$
• D

  $\left(\mathrm{1,5},7\right)$

Solution

$Q$ is reflection of $P$ about line $L$

Coordinate of point $R$ are $\left(2\lambda +\mathrm{1,3}\lambda -\mathrm{2,6}\lambda +7\right)$
$\overrightarrow{PR}=⟨2\lambda -\mathrm{1,3}\lambda +\mathrm{3,6}\lambda +7⟩$
From diagram
$\overrightarrow{PR}\cdot \left(2\widehat{i}+3\widehat{j}+6\widehat{k}\right)=0$
$4\lambda -2+9\lambda +9+36\lambda +42=0$
$49\lambda +49=0\Rightarrow \boxed{\lambda =-1}$
So, coordinates of $R$ are $\left(-1,-5,1\right)$
$\Rightarrow$ coordinates of $Q$ are $\left(-4,-5,2\right)$

Hence, the correct answer is option: 'A'.