The distance of the plane $2x - 3y + 6z + 14 = 0$ from the origin is:

$O$ is the origin and $A$ is $(a, b, c)$. Find the direction cosines of the line $OA$ and the equation of the plane passing through $A$ at a right angle to $OA$.

The plane passing through the point $(4, -1, 2)$ and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{-1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point

The equation of the plane which bisects the line joining $(2, 3, 4)$ and $(6, 7, 8)$ is

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

$A$ plane ( $\pi$ ) passing through the point $(1, 2, -3)$ is perpendicular to the planes $x + y - z + 4 = 0$ and $2x - y + z + 1 = 0$. If the equation of the plane ( $\pi$ ) is $ax + by + cz + 1 = 0$,then $a^2 + b^2 + c^2 =$