If $\alpha$ and $\beta$ are scalars and $\vec{r} = (2+\alpha-3\beta) \hat{i} + (\beta-3) \hat{j} + (2\alpha-5\beta-1) \hat{k}$ is the equation of a plane, then its equation in Cartesian form is:

The equation of the plane,passing through the midpoint of the line segment joining the points $P(1, 2, 5)$ and $Q(3, 4, 3)$ and perpendicular to it,is

Let $P$ be the plane passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of $P$ from the point $(3, -2, 2)$ is

If $(x, y, z)$ is an arbitrary point lying on a plane $P$ which passes through the points $(42, 0, 0)$,$(0, 42, 0)$,and $(0, 0, 42)$,then the value of the expression $3 + \frac{x-11}{(y-19)^{2}(z-12)^{2}} + \frac{y-19}{(x-11)^{2}(z-12)^{2}} + \frac{z-12}{(x-11)^{2}(y-19)^{2}} - \frac{x+y+z}{14(x-11)(y-19)(z-12)}$ is:

$A$ plane passes through $(2,1,2)$ and $(1,2,1)$ and is parallel to the line $2x = 3y$ and $z = 1$. Then the plane also passes through which of the following points?

The distance between the planes given by $r \cdot (i + 2j - 2k) + 5 = 0$ and $r \cdot (i + 2j - 2k) - 8 = 0$ is