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

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 $\sqrt{p^2+q^2+r^2}=$

Given points $A(3, 2, -1)$ and $B(1, 4, 3)$,find the equation of the plane that bisects the segment $AB$ perpendicularly.

$A$ plane meets the coordinate axes at $A, B, C$ respectively such that the centroid of the $\triangle ABC$ is $(2, 3, 5)$. Then,the equation of that plane is

The equation of the plane passing through $(1, -1, 2)$ and perpendicular to the planes $x + 2y - 2z = 4$ and $3x + 2y + z = 6$ is:

$A$ plane $x$ passes through the point $(1, 1, 1)$. If $b, c, a$ are the direction ratios of a normal to the plane,where $a, b, c$ $(a < b < c)$ are the factors of $2001$,then the equation of the plane is