$A$ line joining the points $(1, 2, 0)$ and $(4, 13, 5)$ is perpendicular to a plane. Then the coefficients of $x, y$ and $z$ in the equation of the plane are respectively

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

In the following case,find the distance of the given point from the corresponding given plane.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

If the points $A(2-x, 2, 2)$,$B(2, 2-y, 2)$,$C(2, 2, 2-z)$,and $D(1, 1, 1)$ are coplanar,then the locus of point $P(x, y, z)$ is

The $x$-intercept of a plane $\pi$ passing through the point $(1,1,1)$ is $\frac{5}{2}$ and the perpendicular distance from the origin to the plane $\pi$ is $\frac{5}{7}$. If the $y$-intercept of the plane $\pi$ is negative and the $z$-intercept is positive,then its $y$-intercept is

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