If a plane cuts the coordinate axes at $A, B$ and $C$ respectively such that the centroid of the triangle $ABC$ is $(6, 6, 3)$, then find the equation of that plane.

If $\alpha$ and $\beta$ are scalars and $\vec{r} = (2+\alpha-3\beta) \hat{i} + (\beta-3) \hat{j} + (2\alpha-5\beta-1) \hat{k}$ is the equation of a plane, then its equation in Cartesian form is:

Find the equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$.

$A$ 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

If the perpendicular distance from $(1, 2, 4)$ to the plane $2x + 2y - z + k = 0$ is $3$,then $k =$

Find the angle between the planes $2x - y + z = 6$ and $x + y + 2z = 3$.