$A$ plane passes through the point $A(2, 1, -3)$. If the distance of this plane from the origin is maximum,then its equation is

The equation of the plane in Cartesian form, which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector drawn from the origin being $2 \hat{i}-3 \hat{j}+4 \hat{k}$, is

The coordinates of the foot of the perpendicular drawn from the origin to the plane $3x + 2y + 6z = 56$ are:

$A$ plane $\pi$ is passing through $(2,0,1)$ and $(3,-3,4)$ and is perpendicular to the plane $x-2y+z=6$. Among the following,the plane that is perpendicular to the plane $\pi$ is:

$P$ is a fixed point $(a, a, a)$ on a line through the origin equally inclined to the axes. Then,any plane through $P$ perpendicular to $OP$ makes intercepts on the axes,the sum of whose reciprocals is equal to:

The direction ratios of the normal to the plane passing through the origin and the line of intersection of the planes $x+2y+3z=4$ and $4x+3y+2z=1$ are . . . . . .