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3d Geometry

JEE Main 2022 (Online) 30th June Morning Shift

MCQ+4 / -12022

The distance of the point (3, 2, \(-\)1) from the plane \(3x - y + 4z + 1 = 0\) along the line \({{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}\) is equal to :

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