Consider a plane x+y−z=1 and point A(1,2,−3). A line L has the equation x=1+3r,y=2−r and z=3+4r.
The distance between the points on the line which are at a distance of 4√3 from the plane is
The correct option is A 4√26
The distance of point (1+3r,2−r,3+4r) from the plane is ∣∣∣1+3r+2−r−3−4r−1√1+1+1∣∣∣=4√3
Hence, the points are A(112,12,182) and B(−132,92,−142)