Let P(-7,1,-5) be a point on a plane and let O be the origin. If OP is normal to the plane, then the equation of the plane is

Let P(-7,1-5) be a point on a plane and let O be the origin If OP is normal to the plane , then the equation on the plane is a) 7x-y+5z+75=0 b) 7x+y-5z+73=0 c) 7x+y+5z+73=0 d) 7x-y-5x+75=0

Let A be the set of all points in a plane and let O be the origin Let R={(p,q):OP=OQ}. then ,R is

Let P be the image of the point (3, 1, 7) with respect to the plane x-y+z=3 . Then, the equation of the plane passing through P and containing the straight line (x)/(1)=(y)/(2)=(z)/(1) is

Let P be the image of the point (3, 1, 7) with respect to the plane x-y+z=3 . Then, the equation of the plane passing through P and containing the straight line (x)/(1)=(y)/(2)=(z)/(1) is

Let P be the image of the point (3, 1, 7) with respect to the plane x -y + z = 3 . Then, the equation of the plane passing through P and containing the straight line (x)/(1) = (y)/(2) = (z)/(1) is

The foot of the perpendicular drawn from the origin to the plane is the point (2, 5, 7). Find the equation of the plane.