$A$ 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 $(2, 3, 4)$ is

$A$ bisector of the angle between the normals of the planes $4x + 3y = 5$ and $x + 2y + 2z = 4$ is along the vector

$A$ 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 direction ratios of the lines $OA$ and $OB$ are $1, -2, -1$ and $3, -2, 3$. Then the direction cosines of the normal of plane $AOB$,where $O$ is the origin,are

Find the equation of the plane passing through the points $(2, 1, 0)$,$(3, -2, -2)$,and $(3, 1, 7)$.

The equation of the plane in normal form passing through the point $A(\vec{a})$, parallel to a vector $\vec{b}$ and containing a vector $\vec{c}$ is