Show that the line `(x+3)/(4)=(y-5)/(-1)=(z+7)/(2)` lies in the plane `x-2y-3z-8=0`.

A point P lies on a line through Q(1,-2,3) and is parallel to the line (x)/(1)=(y)/(4)=(z)/(5) . If P lies on the plane 2x+3y-4z+22=0 , then segment PQ equal to

Find the equation of the projection of the line (x-1)/(2)=(y+1)/(-1)=(z-3)/(4) on the plane x+2y+z=9 .

The coordinates of the point where the line (x-3)/(1)=(y-4)/(2)=(z-5)/(2) meets the plane x+y+z=17 are -

The point in which the line (x-2)/(3)=(y+1)/(4)=(z-2)/(12) meets the plane x-2y+z=20 is __

Find the angle between the line (x-1)/(3)=(y+1)/(2)=(z+2)/(4) and the plane 2x+y-3z+4=0 .

The image of the line (x -1)/(3) = (y -3)/(1) = (z -4)/(-5) in the plane 2x - y + z + 3=0 is the line-

The value k for which the line (x-1)/(8)=(y+2)/(k)=(z-3)/(-6) is perpendicular to the plane 4x+2y-3z=5 is -

The angle between the line (x-1)/(2)=(y-2)/(1)=(z+3)/(-1) and the plane x+y+4=0 is __

Find the value of lambda such that the line (x-2)/(6)=(y-1)/( lambda)=(z+5)/(-4) is prependicular to the plane 3x-y-2z=7

The projection of the line (x+1)/(-1)=(y)/(2)=(z-1)/(3) on the plane x-2y+z=6 is the line of intersection of this plane with the plane