What is the equation of the plane passing through the point $(1, 1, 0)$ and perpendicular to the line $\vec{r} = (2, 3, 4) + k(3, 4, 5)$,where $k \in R$?

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 equation of a plane passing through $A(1, p, 2)$ and $B(3, 2, 4)$ and parallel to the $z$-axis is $3x - 2y - q = 0$, then:

If the plane $56x + 4y + 9z = 2016$ meets the coordinate axes at points $A$,$B$,and $C$,then the centroid of the $\triangle ABC$ is

Let the plane $\pi$ pass through the point $(1,0,1)$ and be perpendicular to the planes $2x+3y-z=2$ and $x-y+2z=1$. Let the equation of the plane passing through the point $(11,7,5)$ and parallel to the plane $\pi$ be $ax+by-z-d=0$. Then,$\frac{a}{b}+\frac{b}{d}=$

The direction cosines of the normal drawn to the plane passing through the points $(2,-1,5)$,$(1,-3,4)$,and $(5,2,1)$ are