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(C) Let the equation of the plane be $a(x - 1) + b(y + 2) + c(z - 1) = 0$,where $\vec{n} = (a, b, c)$ is the normal vector to the plane.Since the plan

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$2\sqrt{2}$

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(C) Let the equation of the plane be $a(x - 1) + b(y + 2) + c(z - 1) = 0$,where $\vec{n} = (a, b, c)$ is the normal vector to the plane.Since the plane is perpendicular to $2x - 2y + z = 0$ and $x - y + 2z = 4$,the normal vector $\vec{n}$ is parallel to the cross product of the normals of the given planes,$\vec{n_1} = (2, -2, 1)$ and $\vec{n_2} = (1, -1, 2)$.$\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 1 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 + 2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.We can take the normal vector as $\vec{n} = (1, 1, 0)$.The equation of the plane is $1(x - 1) + 1(y + 2) + 0(z - 1) = 0$,which simplifies to $x + y + 1 = 0$.The distance of the point $(1, 2, 2)$ from the plane $x + y + 1 = 0$ is given by $d = \frac{|1 + 2 + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$ units.

$A$ plane passes through $(1, -2, 1)$ and is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. The distance of the point $(1, 2, 2)$ from this plane is . . . . . . units.

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