The distance between the line frac{x - 1}{3} | vedclass

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(D) First,check if the line is parallel to the plane. The direction vector of the line is $\vec{v} = (3, -2, 2)$ and the normal to the plane is $\vec{

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$3$

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(D) First,check if the line is parallel to the plane. The direction vector of the line is $\vec{v} = (3, -2, 2)$ and the normal to the plane is $\vec{n} = (2, 2, -1)$.Since $\vec{v} \cdot \vec{n} = (3)(2) + (-2)(2) + (2)(-1) = 6 - 4 - 2 = 0$,the line is parallel to the plane.To find the distance,pick any point on the line,such as $P(1, -2, 1)$.The perpendicular distance $d$ from point $(x_1, y_1, z_1)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.Substituting the values: $d = \frac{|2(1) + 2(-2) - 1(1) - 6|}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{|2 - 4 - 1 - 6|}{\sqrt{4 + 4 + 1}} = \frac{|-9|}{\sqrt{9}} = \frac{9}{3} = 3$.

The distance between the line $\frac{x - 1}{3} = \frac{y + 2}{-2} = \frac{z - 1}{2}$ and the plane $2x + 2y - z = 6$ is

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