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 $P$ to the plane is

$\text{Given: } P(1,\ -2,\ 1)$ $\text{Plane: } x + 2y - 2z = \alpha, \quad \alpha > 0$ $\text{Distance of } P \text{ from the plane} = 5$ $\text{To find: foot of the perpendicular from } P \text{ to the plane}$ $\text{Plane } \pi: x + 2y - 2z = \alpha, \quad P(1,-2,1)$ $\text{Plan: } 1.\ \text{find } \alpha \qquad 2.\ F = P - \dfrac{(n\cdot P - d)}{|n|^2}\,n$ $\text{Distance} = \dfrac{|ax_1 + by_1 + cz_1 - \alpha|}{\sqrt{a^2 + b^2 + c^2}}$ $5 = \dfrac{|1(1) + 2(-2) - 2(1) - \alpha|}{\sqrt{1 + 4 + 4}} = \dfrac{|-5 - \alpha|}{3}$ $5 = \dfrac{5 + \alpha}{3} \ \Rightarrow \ 5 + \alpha = 15 \ \Rightarrow \ \alpha = 10$ $\pi: x + 2y - 2z = 10, \quad n = (1,2,-2), \quad d = 10$ $n\cdot P = 1(1) + 2(-2) + (-2)(1) = 1 - 4 - 2 = -5$ $n\cdot P - d = -5 - 10 = -15$ $|n|^2 = 1 + 4 + 4 = 9$ $F = P - \dfrac{n\cdot P - d}{|n|^2}\,n = (1,-2,1) - \dfrac{-15}{9}(1,2,-2)$ $= (1,-2,1) + \dfrac{5}{3}(1,2,-2)$ $F = \left(1 + \dfrac{5}{3},\ -2 + \dfrac{10}{3},\ 1 - \dfrac{10}{3}\right)$ $F = \left(\dfrac{8}{3},\ \dfrac{4}{3},\ -\dfrac{7}{3}\right)$ $\dfrac{8}{3} + 2\cdot\dfrac{4}{3} - 2\cdot\left(-\dfrac{7}{3}\right) = \dfrac{8 + 8 + 14}{3} = \dfrac{30}{3} = 10 = d \ \checkmark$ $F = \left(\dfrac{8}{3},\ \dfrac{4}{3},\ -\dfrac{7}{3}\right)$ $\therefore \text{Option } (1)$