The foot of perpendicular from (-1,2 , 3 ) on the plane passing through the points ( 1,- 1, 1 )...

Find the locus of the foot of perpendicular from the point (2,1) on the variable line passing through the point (0,0).

The foot of the perpendicular drawn form the point (4,2,3) to the line joining the points (1,-2,3) and (1,1,0) lies on the plane:

The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1) having normal perpendicular to both the lines (x-1)/(1)=(y+2)/(-2)=(z-4)/(3) and (x-2)/(2)=(y+1)/(-1)=(z+7)/(-1) is

Equation of the plane passing through the points (1,0,1), (-1,1,1) and (2,3,1) is

Find the unit vector perpendicular to the plane of a triangle passing through the points (1,1,2),(2,3,5) and (1,5,5)

If the foot of perpendicular drawn from the point (2, 5, 1) on a line passing through (alpha, 2alpha, 5)" is "((1)/(5),(2)/(5),(3)/(5)) , then alpha is equal to

The locus of the foot of the perpendicular from the origin to a straight line passing through (1,1) is

The equation of the plane passing through the points (1,2,3), (-1,4,2) and (3,1,1) is

The foot of the perpendicular drawn from the origin to a plane is the point (1, -3, 1) . What is the intercept cut on the x-axis by the plane ?