Prove that the planes `12x-15y+16z-28=0, 6x + 6y -7z- 8= 0` and `2x + 35y-39z + 12 = 0` have a common line of intersection.

Consider three planes : 2x+py+6z=8, x+2y+qz=5 and x+y+3z=4 Q. Three planes do not have any common point of intersection if :

Prove that the lines 3x + 2y + z – 5 = 0 = x + y – 2z – 3 and 2x – y – z = 0 = 7x + 10y – 8z – 15 are perpendicular

If 7x + 6y - 2z = 0, 3x + 4y + 2z = 0, x - 2y - 6z = 0 then which option is correct

Show that the planes 2x-y+6z=5 and 5x-2.5y+15z=12 are parallel.

Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

The plane x + 3y + 13 = 0 passes through the line of intersection of the planes 2x - 8y + 4z = p and 3x - 5y + 4z + 10 = 0 If the plane is perpendicular to the plane , 3x - y - 2z - 4 = 0 then the value of p is equal to

The plane x + 3y + 13 = 0 passes through the line of intersection of the planes 2x - 8y + 4z = p and 3x - 5y + 4z + 10 =0 . If the plane is perpendicular to the plane 3x - y - 2z - 4 = 0 , then the value of p is equal to

I. 5x^(2) + 28x + 15 = 0 II. 6y^(2) + 35y + 25 = 0

Are the planes x+y-2z+4=0 and 3x+3y-6z+5=0 intersecting