If the projection of the line x2=y−12=z−11 on a plane P is x1=y−11=z−1−1. Then the distance of plane P from origin is

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$If the projection of the line \frac{x}{2} = \frac{y - 1}{2} = \frac{z - 1}{1} on a plane P is \frac{x}{1} = \frac{y - 1}{1} = \frac{z - 1}{- 1} . Then the distance of plane P from origin is$

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$\frac{2}{\sqrt{3}}$

$\sqrt{6}$

$\sqrt{3}$

$\sqrt{\frac{3}{2}}$

answer is B.

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Detailed Solution

$Projection of line \frac{x}{2} = \frac{y - 1}{2} = \frac{z - 1}{1} on a plane P is$

$\frac{x}{1} = \frac{y - 1}{1} = \frac{z - 1}{- 1}$

Plane through these lines is perpendicular to the plane P Normal to the plane determined by the given lines is
$|\begin{array}{rrr} \hat{i} \hat{j} \hat{k} \\ 2 2 1 \\ 1 1 - 1 \end{array}| = - 3 \hat{i} + 3 \hat{j}$