If the distance of the point $P(1,-2,1)$ from the plane $x+2 y-2 z=\alpha$, where $\alpha>0$, is 5 , then the foot of the perpendicular from $P$ to the plane is

• Correct answer
  A

  $\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$
• B

  $\left(\frac{4}{3},-\frac{4}{3}, \frac{1}{3}\right)$
• C

  $\left(\frac{1}{3}, \frac{2}{3}, \frac{10}{3}\right)$
• D

  $\left(\frac{2}{3},-\frac{1}{3}, \frac{5}{2}\right)$

Solution

$$ \text { Distance of point } P \text { from plane }=5 $$ $$ \begin{gathered} \therefore=\left|\frac{1-4-2-\alpha}{3}\right| \\ \alpha=10 \end{gathered} $$
Foot of perpendicular $$ \begin{aligned} \quad \frac{x-1}{1} & =\frac{y+2}{2}=\frac{z-1}{-2}=\frac{5}{3} \\ \Rightarrow \quad \quad \quad x & =\frac{8}{3}, y=\frac{4}{3}, z=-\frac{7}{3} \end{aligned} $$ Thus, the foot of the perpendicular is $$ A\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right) $$

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