The coordinates of the foot of the perpendicular drawn from the origin to a plane is $(2, 4, -3)$. The equation of the plane is

The angle between a normal to the plane $2x - y + 2z - 1 = 0$ and the $X$-axis is

Let $(\alpha, \beta, \gamma)$ be the image of the point $P (2, 3, 5)$ in the plane $2x + y - 3z = 6$. Then $\alpha + \beta + \gamma$ is equal to

$A$ 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 distance of the point $(5,3,-1)$ from the plane passing through the points $A(2,1,0)$,$B(3,-2,4)$,and $C(1,-3,3)$ is:

$A$ plane meets the coordinate axes at points $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, r, r^2)$. Find the equation of the plane.