Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $3y + 4z - 6 = 0$.

The plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3$ meets the coordinate axes at points $A, B,$ and $C$. The centroid of the triangle $ABC$ is:

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

The distance of the point $(5,3,-1)$ from the plane passing through the points $A(2,1,0)$,$B(3,-2,4)$,and $C(1,-3,3)$ is:

$A$ variable plane passes through a fixed point $(3, 2, 1)$ and meets the $x, y,$ and $z$ axes at $A, B,$ and $C$ respectively. $A$ plane is drawn parallel to the $yz$-plane through $A$,a second plane is drawn parallel to the $zx$-plane through $B$,and a third plane is drawn parallel to the $xy$-plane through $C$. Then,the locus of the point of intersection of these three planes is:

If the angle between the planes $x-2y+3z-5=0$ and $x+\alpha y+2z+7=0$ is $\cos^{-1}\left(\frac{1}{14}\right)$,then the difference between the values of $\alpha$ is