$A$ point on the plane that passes through the points $(1, -1, 6)$, $(0, 0, 7)$ and is perpendicular to the plane $x - 2y + z = 6$ is

If ${P_1}$ and ${P_2}$ are the lengths of the perpendiculars from the points $(2, 3, 4)$ and $(1, 1, 4)$ respectively to the plane $3x - 6y + 2z + 11 = 0$,then ${P_1}$ and ${P_2}$ are the roots of the equation:

If the foot of the perpendicular drawn from $(0,0,0)$ to a plane is $(1,2,3)$,then the equation of the plane is:

The equation of the plane passing through $(-1, 1, 2)$ whose normal makes equal acute angles with the coordinate axes 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

If planes $x - c y - b z = 0$,$c x - y + a z = 0$ and $b x + a y - z = 0$ pass through a straight line then $a^2 + b^2 + c^2 =$