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JEE Main202126 Aug 2021Morning ShiftMathematicsThree Dimensional GeometryActual

Let the line L be the projection of the line x - 1 2 = y - 3 1 = z - 4 2 in the plane x - 2 y - z = 3 . If d is the distance of the point ( 0 , 0 , 6 ) from L , then d 2 is equal to

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Step-by-step solution

Let x - 1 2 = y - 3 1 = z - 4 2 = r Any point on the line P 2 r + 1 ,   r + 3 ,   2 r + 4 For the point P 2 r + 1 ,   r + 3 ,   2 r + 4 to lie on the plane, its co-ordinates should satisfy the equation x - 2 y - z = 3 . ⇒ 2 r + 1 - 2 r + 3 - 2 r + 4 = 3 ⇒ - 2 r - 9 = 3 ⇒ r = - 6 ∴ P - 11 ,   - 3 ,   - 8 Now, let Q h ,   k ,   l be the coordinates of the foot of the perpendicular drawn from 1 ,   3 ,   4 to the plane x - 2 y - z = 3 . ⇒

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