If the equation of the plane which is at a distance of $\frac{1}{3}$ units from the origin and perpendicular to a line whose directional ratios are $(1, 2, 2)$ is $x+py+qz+r=0$,then $\sqrt{p^2+q^2+r^2}=$

If the distance of the point $(1, 1, 1)$ from the origin is half its distance from the plane $x + y + z + k = 0$,then $k = $

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

$A$ plane meets the coordinate axes at $A, B, C$ and $(\alpha, \beta, \gamma)$ is the centroid of the triangle $ABC$. Then the equation of the plane is

Let $\pi$ be the plane passing through the point $(3,-3,1)$ and perpendicular to the line joining the points $(3,4,-1)$ and $(2,-1,5)$. If the equation of the plane containing the points $(3,4,-1),(-1,2,5)$ and perpendicular to the plane $\pi$ is $ax+y+cz-d=0$,then $3(a+c)=$

If $A=(1,8,4)$ and $B=(2,-3,1)$,then the direction cosines of a normal to the plane $AOB$ are