If the foot of perpendicular drawn from the point $\left(\mathrm{2,5},1\right)$ on a line passing through $\left(\alpha ,2\alpha ,5\right)$ is $\left(\frac{1}{5},\frac{2}{5},\frac{3}{5}\right)$, then $\alpha$ is equal to

• A

  $\frac{19}{9}$
• B

  $\frac{11}{570}$
• C

  $\frac{57}{54}$
• Correct answer
  D

  $\frac{1}{25}$

Solution

From diagram $\overrightarrow{PQ}\cdot \overrightarrow{QR}=0$
$\left[\left(2-\frac{1}{5}\right)\hat{i}+\left(5-\frac{2}{5}\right)\hat{j}+\left(1-\frac{3}{5}\right)\hat{k}\right]·\left[\left(\alpha -\frac{1}{5}\right)\hat{i}+\left(2\alpha -\frac{2}{5}\right)\hat{j}+\left(5-\frac{3}{5}\right)\hat{k}\right]=0$
$\Rightarrow \frac{9\alpha }{5}-\frac{9}{25}+\frac{46\alpha }{5}-\frac{46}{25}+\frac{44}{25}=0$
$\Rightarrow \alpha =\frac{1}{25}$

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