The minimum distance $d_{\min}$ from the origin $(0, 0, 0)$ to the plane $2x - y + 2z = 9$ in three-dimensional Cartesian space is:

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### 1. Strategy & Theoretical Principles The shortest perpendicular distance from a point $(x_0, y_0, z_0)$ to a plane $Ax + By + Cz + D = 0$ is: $$d = \frac{|A x_0 + B y_0 + C z_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$ Alternatively, minimizing $f(x, y, z) = x^2 + y^2 + z^2$ subject to $g(x, y, z) = 2x - y + 2z - 9 = 0$ via Lagrange multipliers $\nabla f = \lambda \nabla g$ leads to the same result. ### 2. Analytical Formulation & Calculation For $(x_0, y_0, z_0) = (0, 0, 0)$ and plane $2x - y + 2z - 9 = 0$: $$A = 2, \quad B = -1, \quad C = 2, \quad D = -9$$ $$d = \frac{|2(0) - 1(0) + 2(0) - 9|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{|-9|}{\sqrt{4 + 1 + 4}} = \frac{9}{\sqrt{9}} = \frac{9}{3} = 3.0$$ ### 3. Distractor Forensics - Omission of square root in denominator gives $9/9 = 1.0$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Remember $\sqrt{A^2 + B^2 + C^2}$ in the denominator. - **Standard Reference:** Thomas, G.B., *Calculus*, 14th Edition, Pearson, Chapter 14.

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