The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x + 6y - 3z = 63$ are

If a plane passes through the point $(1, 1, 1)$ and is perpendicular to the line $\frac{x - 1}{3} = \frac{y - 1}{0} = \frac{z - 1}{4}$,then its perpendicular distance from the origin is

In space,the equation $by + cz + d = 0$ represents a plane perpendicular to the

$A$ plane $\pi$ passing through the point $(1,1,1)$ is perpendicular to the line joining the points $(6,3,2)$ and $(1,-4,-9)$. If $ax+by+cz-23=0$ is the equation of the plane $\pi$, then $a+b-c=$

$P$ 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:

Let $\alpha, \beta, \gamma, \delta$ be real numbers such that $\alpha^2+\beta^2+\gamma^2 \neq 0$ and $\alpha+\gamma=1$. Suppose the point $(3,2,-1)$ is the mirror image of the point $(1,0,-1)$ with respect to the plane $\alpha x+\beta y+\gamma z=\delta$. Then which of the following statements is/are $TRUE$?
$(A)$ $\alpha+\beta=2$
$(B)$ $\delta-\gamma=3$
$(C)$ $\delta+\beta=4$
$(D)$ $\alpha+\beta+\gamma=\delta$