Equation of line passing through the point (2, 3, 1) and parallel to the line of intersection of the planes x - 2y - z + 5 = 0 and x + y + 3z = 6 is
D.
Let the DR's of !me passmg through intersectron of the given two planes x - 2y - z + 5 = 0 and , x + y + 3z = 6 are a, b and c
Then, a + 2b - c = 0
and a + b + 3c = 0
Since, required line passes through (2, 3, 1) and parallel to above line Hence, its equation will be
Foot of perpendicular drawn from the origin to the plane 2x - 3y + 4z = 29 is
(7, - 1, 3)
(5, - 1, 4)
(5, - 2, 3)
(2, - 3, 4)
The vector equation of the plane, which is at a distance of , from the origin and the normal from the origin is is
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 5y + 8 = 0.