The length of the perpendicular (in units) from the point $\left(\mathrm{1,2},4\right)$ on the straight line $\frac{x-2}{1}=\frac{y-7}{2}=\frac{z-3}{-1}$ lies in the interval

• A

  $\left(1,\frac{3}{2}\right)$
• B

  $\left(\mathrm{2,3}\right)$
• Correct answer
  C

  $\left(\mathrm{0,2}\right]$
• D

  $\left[\mathrm{4,5}\right)$

Solution

Let, point $A$ is $\left(\mathrm{1,2},4\right)$ and any general point on the line is $B\left(\lambda +\mathrm{2,2}\lambda +7,-\lambda +3\right)$, 
$\overrightarrow{AB}=\left(\lambda +1\right)\widehat{i}+\left(2\lambda +5\right)\widehat{j}+\left(-\lambda -1\right)\widehat{k}$
and a vector parallel to the line is $\overrightarrow{b}=\widehat{i}+2\widehat{j}-\widehat{k}$
$\overrightarrow{AB}\cdot \overrightarrow{b}=0\Rightarrow \lambda +1+4\lambda +10+\lambda +1=0$
$\Rightarrow \lambda =-2$
The length of the perpendicular $=\left|\overrightarrow{AB}\right|$
$=\sqrt{{\left(-2+1\right)}^2+{\left(-4+5\right)}^2+{\left(2-1\right)}^2}=\sqrt{3}$ units

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