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

$P(1, -2, 1), \quad \text{plane } x + 2y - 2z = \alpha \ (\alpha > 0)$ $\text{distance of } P \text{ from the plane} = 5$ $\text{To find: foot of the perpendicular from } P \text{ to the plane}$ $P(1, -2, 1), \quad \text{plane } x + 2y - 2z = \alpha$ $\vec{n} = (1, 2, -2), \quad |\vec{n}| = \sqrt{1 + 4 + 4} = 3$ $d = \dfrac{|1(1) + 2(-2) - 2(1) - \alpha|}{3} = \dfrac{|-5 - \alpha|}{3}$ $\dfrac{|-5 - \alpha|}{3} = 5 \Rightarrow |-5 - \alpha| = 15$ $\alpha > 0 \Rightarrow -5 - \alpha = -15 \Rightarrow \alpha = 10$ $\text{Plane: } x + 2y - 2z = 10$ $\text{Foot} = P - \dfrac{\vec{n}\cdot P - d}{|\vec{n}|^2}\,\vec{n}$ $|\vec{n}|^2 = 9$ $\vec{n}\cdot P = 1 - 4 - 2 = -5$ $\vec{n}\cdot P - d = -5 - 10 = -15$ $\text{Foot} = P - \dfrac{-15}{9}(1, 2, -2) = P + \dfrac{5}{3}(1, 2, -2)$ $= \left(1 + \dfrac{5}{3},\ -2 + \dfrac{10}{3},\ 1 - \dfrac{10}{3}\right) = \left(\dfrac{8}{3}, \dfrac{4}{3}, -\dfrac{7}{3}\right)$ $\text{Check: } \dfrac{8}{3} + 2\cdot\dfrac{4}{3} - 2\left(-\dfrac{7}{3}\right) = \dfrac{30}{3} = 10\ \checkmark$ $\boxed{\left(\dfrac{8}{3}, \dfrac{4}{3}, -\dfrac{7}{3}\right)} \quad \text{option (1)}$