$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$. Then,the equation of the plane is

If the distance between the plane,$23x - 10y - 2z + 48 = 0$ and the plane containing the lines $\frac{x+1}{2} = \frac{y-3}{4} = \frac{z+1}{3}$ and $\frac{x+3}{2} = \frac{y+2}{6} = \frac{z-1}{\lambda}$ $(\lambda \in R)$ is equal to $\frac{k}{\sqrt{633}}$,then $k$ is equal to

$A$ 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 equation of the plane passing through a point $A(2, -1, 3)$ and parallel to the vectors $\vec{a} = (3, 0, -1)$ and $\vec{b} = (-3, 2, 2)$ is:

The equation of the plane passing through the point $(1, -3, -2)$ and perpendicular to the planes $x + 2y + 2z = 5$ and $3x + 3y + 2z = 8$ is:

The perpendicular distance of the origin from the plane $2x + y - 2z - 18 = 0$ is (in $\text{ units}$)