The perpendicular distance of the point (2, 4, -1) from the line x+51=y+34=6−z9 is
7
14
Distance of the point →c from the line →r= →a+t→b is |→b×(→a−→c)||→b|
Therefore distance = 7 units
Find the distance of a point (2,4,-1) from the line. x+51=y+34=z−6−9.