$A$ plane passes through the point $A(2, 1, -3).$ If the distance of this plane from the origin is maximum,then its equation is

The Cartesian equation of the plane,passing through the points $(3,1,1)$,$(1,2,3)$ and $(-1,4,2)$,is

$A$ plane meets the coordinate axes at points $A$, $B$ and $C$, such that the centroid of triangle $ABC$ is $(2, -\frac{2}{3}, \frac{1}{2})$. The perpendicular distance from the origin to this plane is:

The distance of the plane $\vec{r} = (\hat{i} - \hat{j}) + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k})$ from the origin is

If the distance between the planes $2x + y + z + 1 = 0$ and $2x + y + z + \alpha = 0$ is $3$ units,then the product of all possible values of $\alpha$ is

Let $\pi$ be the plane that passes through the point $(-2, 1, -1)$ and is parallel to the plane $2x - y + 2z = 0$. Then the foot of the perpendicular drawn from the point $(1, 2, 1)$ to the plane $\pi$ is