If a plane passes through (1, -2, 1) and is p | vedclass

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(C) The normal vector $\vec{n}$ of the required plane is perpendicular to the normal vectors $\vec{n}_1 = (2, -2, 1)$ and $\vec{n}_2 = (1, -1, 2)$ of

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$2\sqrt{2}$

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(C) The normal vector $\vec{n}$ of the required plane is perpendicular to the normal vectors $\vec{n}_1 = (2, -2, 1)$ and $\vec{n}_2 = (1, -1, 2)$ of the given planes.Thus,$\vec{n} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 1 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 + 2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.We can take the normal vector as $\vec{n} = (1, 1, 0)$.The equation of the plane passing through $(1, -2, 1)$ is $1(x - 1) + 1(y + 2) + 0(z - 1) = 0$,which simplifies to $x + y + 1 = 0$.The distance of the plane $x + y + 1 = 0$ from the point $(1, 2, 2)$ is given by $d = \frac{|1(1) + 1(2) + 0(2) + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|1 + 2 + 1|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.

If a plane passes through $(1, -2, 1)$ and is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,then the distance of that plane from the point $(1, 2, 2)$ is

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