The equation of the plane whose intercepts on $x, y, z$ axes are $1, 2, 4$ respectively 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:

If the given planes $ax + by + cz + d = 0$ and $a'x + b'y + c'z + d' = 0$ are mutually perpendicular,then:

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the $X$-axis at $A$,$Y$-axis at $B$,and $Z$-axis at $C$. The area of $\triangle ABC$ is:

If the foot of the perpendicular drawn from the point $P(2,0,-3)$ to the plane $\pi$ is $F(1,-2,0)$ and the equation of the plane $\pi$ is $ax+by-3z+d=0$, then $a+b+d=$

$A$ plane meets the coordinate axes at the points $A, B, C$ respectively in such a way that the centroid of $\triangle ABC$ is $(1, r, r^2)$ for some real $r$. If the plane passes through the point $(5, 5, -12)$, then $r=$