Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $5y + 8 = 0$.

$A$ plane $\pi$ is passing through $(2,0,1)$ and $(3,-3,4)$ and is perpendicular to the plane $x-2y+z=6$. Among the following,the plane that is perpendicular to the plane $\pi$ is:

The equation of the plane which bisects the angle between the planes $3x - 6y + 2z + 5 = 0$ and $4x - 12y + 3z - 3 = 0$ which contains the origin is

$A$ plane passes through a fixed point $A(1, -2, 3)$. If the locus of the foot of the perpendicular to it from the point $B(0, 1, 2)$ is $x^2 + y^2 + z^2 + \alpha x + \beta y + \gamma z + \delta = 0$,then the value of $\alpha + \beta + \gamma + \delta$ is

The equations of planes parallel to the plane $x+2y+2z+8=0$,which are at a distance of $2$ units from the point $(1,1,2)$ are

Find the equation of the plane passing through the points $(2, 2, 1)$ and $(9, 3, 6)$ and perpendicular to the plane $2x + 6y + 6z - 1 = 0$.