What is the distance between the line frac{x | vedclass

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(D) The line is given by $\frac{x - 1}{3} = \frac{y + 2}{-2} = \frac{z - 1}{2}$. The direction vector of the line is $\vec{b} = 3\hat{i} - 2\hat{j} +

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$3$

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(D) The line is given by $\frac{x - 1}{3} = \frac{y + 2}{-2} = \frac{z - 1}{2}$. The direction vector of the line is $\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}$.The normal vector to the plane $2x + 2y - z = 6$ is $\vec{n} = 2\hat{i} + 2\hat{j} - 1\hat{k}$.Check if the line is parallel to the plane: $\vec{b} \cdot \vec{n} = (3)(2) + (-2)(2) + (2)(-1) = 6 - 4 - 2 = 0$. Since the dot product is $0$,the line is parallel to the plane.Take a point $P(1, -2, 1)$ on the line. The distance $d$ from point $P$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.Substituting the values: $d = \frac{|2(1) + 2(-2) - 1(1) - 6|}{\sqrt{2^2 + 2^2 + (-1)^2}} = \frac{|2 - 4 - 1 - 6|}{\sqrt{4 + 4 + 1}} = \frac{|-9|}{\sqrt{9}} = \frac{9}{3} = 3$.

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

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