The perpendicular distance of the point $(1, -1, 2)$ from the plane $x + 2y + z = 4$ is

$A$ plane $\pi$ given by $ax + by + 11z + d = 0$ is perpendicular to the planes $2x - 3y + z = 4$ and $3x + y - z = 5$. The perpendicular distance from the origin to the plane $\pi$ is $\sqrt{6}$ units. If all the intercepts made by the plane $\pi$ on the coordinate axes are positive,then $d =$

$A$ plane passing through $(-1, 2, 3)$ and whose normal makes equal angles with the coordinate axes is

$A$ plane is parallel to two lines,whose direction ratios are $1, 0, -1$ and $-1, 1, 0$ and it contains the point $(1, 1, 1)$. If it cuts coordinate axes ($X, Y, Z$-axes respectively) at $A, B, C$,then the volume of the tetrahedron $OABC$ is (in cubic units):

The distance between two parallel planes $2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$ is:

If a plane passes through the point $(1, 1, 1)$ and is perpendicular to the line $\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}$,then its perpendicular distance from the origin is