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(B) The equation of a plane parallel to $x - 2y + 2z - 5 = 0$ is of the form $x - 2y + 2z + k = 0$.Given that the distance from the origin $(0, 0, 0)$

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$x - 2y + 2z - 3 = 0$

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(B) The equation of a plane parallel to $x - 2y + 2z - 5 = 0$ is of the form $x - 2y + 2z + k = 0$.Given that the distance from the origin $(0, 0, 0)$ to this plane is $1$ unit.The formula for the distance from a point $(x_1, y_1, z_1)$ to a plane $Ax + By + Cz + D = 0$ is $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.Substituting the values,we get $1 = \frac{|1(0) - 2(0) + 2(0) + k|}{\sqrt{1^2 + (-2)^2 + 2^2}}$.$1 = \frac{|k|}{\sqrt{1 + 4 + 4}} = \frac{|k|}{\sqrt{9}} = \frac{|k|}{3}$.Therefore,$|k| = 3$,which implies $k = 3$ or $k = -3$.Substituting these values back into the equation,we get $x - 2y + 2z + 3 = 0$ or $x - 2y + 2z - 3 = 0$.Comparing with the given options,$x - 2y + 2z - 3 = 0$ is the correct choice.

Find the equation of a plane that is at a unit distance from the origin and is parallel to the plane $x - 2y + 2z - 5 = 0$.

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