The direction ratios of the normal to the plane passing through $(0,0,1)$,$(0,1,2)$,and $(1,0,3)$ are:

Find the distance of the point $(3, -2, 1)$ from the plane $2x - y + 2z + 3 = 0$. (in $/3$)

The foot of the perpendicular drawn from the origin to the plane is $(4, -2, -5)$. Hence,the equation of the plane is

Find the equation of a plane which is at a distance of $3 \sqrt{3}$ units from the origin and the normal to which is equally inclined to the coordinate axes.

$A$ point moves such that its distances from the points $(3, 4, -2)$ and $(2, 3, -3)$ are equal. What is the locus of the point?

$A$ plane $\pi$ given by $ax + by + 11z + d = 0$ is perpendicular to the planes $2x - 3y + z = 4$ and $3x + y - z = 5$. The perpendicular distance from the origin to the plane $\pi$ is $\sqrt{6}$ units. If all the intercepts made by the plane $\pi$ on the coordinate axes are positive,then $d =$