Question

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

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

Answer 1

The correct option is A

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

Distance from plane
$15=\mid \alpha +5\mid$
Let $(x_1,y_1,z_1)$ be the foot of perpendicular.
Then, $\frac{x_1-1}{1}=\frac{y_1+2}{2}=\frac{z_1-1}{-2}=\frac{-(-15)}{9}$
$(x_1,y_1,z_1)=(\frac{8}{3},\frac{4}{3},\frac{-7}{3})$