Consider a plane `x+2y+3z=15` and a line `(x-1)/(2)=(y+1)/(3)=(z-2)/(4)` then find the distance of origin from point of intersection of line and plane.

The plane x-2y+z=6 and line (x)/(1)=(y)/(2)=(z)/(3) are related as the line :

Consider a plane P:2x+y-z=5 , a line L:(x-3)/(2)=(y+1)/(-3)=(z-2)/(-1) and a point A(3, -4,1) . If the line L intersects plane P at B and the xy plane at C, then the area (in sq. units) of DeltaABC is

If the line (x-1)/(2)=(y+1)/(3)=(z-2)/(4) meets the plane, x+2y+3z=15 at a point P, then the distance of P from the origin is

Line (x-2)/(-2) = (y+1)/(-3) = (z-5)/1 intersects the plane 3x+4y+z = 3 in the point

The line (x)/(1)=(y)/(2)=(z)/(2) and plane 2x-y+z=2 cut at a point A .The distance of point A from origin is:

The distance of the point (1,1,9) from the point of intersection of the line (x-3)/1=(y-4)/2 = (z-5)/2 and plane x+y+z=17

Find the distance between the line (x+1)/(-3)=(y-3)/(2)=(z-2)/(1) and the plane x+y+z+3=0