The Cartesian equation of the plane passing through the point $A(7, 8, 6)$ and parallel to the $XY$-plane is:

Let the plane $ax + by + cz = d$ pass through $(2, 3, -5)$ and be perpendicular to the planes $2x + y - 5z = 10$ and $3x + 5y - 7z = 12$. If $a, b, c, d$ are integers,$d > 0$,and $\text{gcd}(|a|, |b|, |c|, d) = 1$,then the value of $a + 7b + c + 20d$ is equal to

$A$ plane which is perpendicular to two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$ passes through $(1, 2, 1)$. The distance of the plane from the point $(2, 3, 4)$ is

$A$ plane $x$ passes through the point $(1, 1, 1)$. If $b, c, a$ are the direction ratios of a normal to the plane,where $a, b, c$ $(a < b < c)$ are the factors of $2001$,then the equation of the plane is

The perpendicular distance from the origin to the plane $x + 2y - 2z + 5 = 0$ equals $.........$ units.

$A$ plane meets the coordinate axes at $A, B, C$ respectively such that the centroid of the $\triangle ABC$ is $(2, 3, 5)$. Then,the equation of that plane is