Let $A=(-3,-2,7)$ and $B=(3,1,-2)$. Let a plane perpendicular to the line segment $AB$ divide $AB$ in the ratio $2:1$. Then the intercept made by the plane on the $y$-axis is

Find the equation of a plane which is at a distance of $3 \sqrt{3}$ units from the origin and the normal to which is equally inclined to the coordinate axes.

The direction ratios of the normal to the plane passing through $(0,0,1)$,$(0,1,2)$,and $(1,0,3)$ are:

The distance between the planes given by $r \cdot (i + 2j - 2k) + 5 = 0$ and $r \cdot (i + 2j - 2k) - 8 = 0$ is

The volume of the tetrahedron formed by the planes whose equations are $y + z = 0, z + x = 0, x + y = 0$ and $x + y + z = 2$ is

$P_1$ and $P_2$ are two distinct and intersecting planes. Three non-collinear points lie on $P_1$ and another three non-collinear points lie on $P_2$ (none being on the line of intersection of the planes). Then the maximum number of tetrahedrons formed using these six points is: