If the foot of the perpendicular drawn from the origin to a plane is $P(2,-1,4)$,then the equation of the plane is

The equation of a plane,containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(2, 1, 0)$,is

The equation of the plane passing through $(1, 2, 3)$ and parallel to the plane $2x + 3y - 4z = 0$ is

$A$ variable plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$,which is at a unit distance from the origin,cuts the coordinate axes at $A, B$,and $C$. If the centroid $(x, y, z)$ of $\triangle ABC$ satisfies $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=k$,then $k$ equals:

Find the vector and cartesian equations of the plane which passes through the point $(5, 2, -4)$ and is perpendicular to the line with direction ratios $2, 3, -1$.

If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin $(0, 0, 0)$ to a plane,then the equation of that plane is: