1. **State the problem:** We are given a point $P(1, -2, 1)$ and a plane defined by the equation $x + 2y - 2z = \alpha$, where $\alpha > 0$. The distance from $P$ to the plane is 5. We need to find the foot of the perpendicular from $P$ to the plane. 2. **Recall the distance formula from a point to a plane:** The distance $d$ from point $(x_0, y_0, z_0)$ to the plane $Ax + By + Cz + D = 0$ is given by $$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$ 3. **Rewrite the plane equation:** Given $x + 2y - 2z = \alpha$, rewrite as $$x + 2y - 2z - \alpha = 0$$ So, $A=1$, $B=2$, $C=-2$, and $D = -\alpha$. 4. **Apply the distance formula:** Substitute $P(1, -2, 1)$ and $d=5$: $$5 = \frac{|1(1) + 2(-2) - 2(1) - \alpha|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|1 - 4 - 2 - \alpha|}{\sqrt{1 + 4 + 4}} = \frac{| -5 - \alpha|}{3}$$ 5. **Solve for $\alpha$:** Multiply both sides by 3: $$15 = | -5 - \alpha|$$ Since $\alpha > 0$, consider two cases: - Case 1: $-5 - \alpha = 15 \implies \alpha = -20$ (not valid since $\alpha > 0$) - Case 2: $-5 - \alpha = -15 \implies \alpha = 10$ So, $\alpha = 10$. 6. **Find the foot of the perpendicular:** The foot of the perpendicular $F(x_f, y_f, z_f)$ lies on the plane and on the line passing through $P$ in the direction of the plane's normal vector $\vec{n} = (1, 2, -2)$. Parametric form of the line: $$x = 1 + t(1), \quad y = -2 + t(2), \quad z = 1 + t(-2)$$ 7. **Substitute into the plane equation:** $$x + 2y - 2z = 10$$ Substitute parametric coordinates: $$(1 + t) + 2(-2 + 2t) - 2(1 - 2t) = 10$$ Simplify: $$1 + t - 4 + 4t - 2 + 4t = 10$$ $$ (t + 4t + 4t) + (1 - 4 - 2) = 10$$ $$9t - 5 = 10$$ $$9t = 15$$ $$t = \frac{15}{9} = \frac{5}{3}$$ 8. **Calculate coordinates of $F$:** $$x_f = 1 + \frac{5}{3} = \frac{8}{3}$$ $$y_f = -2 + 2 \times \frac{5}{3} = -2 + \frac{10}{3} = \frac{4}{3}$$ $$z_f = 1 - 2 \times \frac{5}{3} = 1 - \frac{10}{3} = -\frac{7}{3}$$ 9. **Final answer:** The foot of the perpendicular is $$\boxed{\left( \frac{8}{3}, \frac{4}{3}, -\frac{7}{3} \right)}$$