Home
Class 12
MATHS
Lying in the plane x+y+z=6 is a line L p...

Lying in the plane `x+y+z=6` is a line L passing through (1, 2, 3) and perpendicular to the line of intersection of planes `x+y+z=6` and `2x-y+z=4`, then the equation of L is

Promotional Banner

Similar Questions

Explore conceptually related problems

The equation of the line passing through (1, 1, 1) and perpendicular to the line of intersection of the planes x+2y-4z=0 and 2x-y+2z=0 is

The equation of the line passing through (1, 1, 1) and perpendicular to the line of intersection of the planes x+2y-4z=0 and 2x-y+2z=0 is

The equation of the line passing through (1, 1, 1) and perpendicular to the line of intersection of the planes x+2y-4z=0 and 2x-y+2z=0 is

The equation of the line passing through (1, 1, 1) and perpendicular to the line of intersection of the planes x+2y-4z=0 and 2x-y+2z=0 is

The equation of plane passing through (2,1,0) and line of intersection of planes x-2y+3z=4 and x-y+z=3 is

Find the equation of line of intersection of the planes 3x-y+ z=1 and x + 4 y -2 z =2.

Find the equation of the plane passing through the line of intersection of the planes .x+ y+ z = 6 and 2x+ 3y+ 4z-5 = 0 and the point (1,1,1).