The distance between the parallel planes $2x - 2y + z + 3 = 0$ and $4x - 4y + 2z + 5 = 0$ is:

The angle between the planes $2x - y + z = 6$ and $x + y + 2z = 3$ is

$A$ plane which is perpendicular to two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,passes through $(1, -2, 1)$. The distance of the plane from the point $(1, 2, 2)$ is

$A$ plane $P$ is parallel to two lines whose direction ratios are $-2, 1, -3$ and $-1, 2, -2$,and it contains the point $(2, 2, -2)$. Let $P$ intersect the coordinate axes at the points $A, B, C$ making the intercepts $\alpha, \beta, \gamma$. If $V$ is the volume of the tetrahedron $OABC$,where $O$ is the origin and $p = \alpha + \beta + \gamma$,then the ordered pair $(V, p)$ is equal to.

The equation of the plane passing through $(-2, 2, 2)$ and $(2, -2, -2)$ and perpendicular to the plane $9x - 13y - 3z = 0$ is

Distance between the two planes: $2x + 3y + 4z = 4$ and $4x + 6y + 8z = 12$ is