The equation of the plane perpendicular to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and passing through the point $(2, 3, 4)$ is:

If a plane passing through the points $(2,3,0), (0,-5,2)$ and $(-2,0,3)$ meets the $X, Y, Z$-axes in $A, B, C$ respectively, then $A=$

The vector equation of the plane passing through the point $A(-2, 7, 5)$ and parallel to the vectors $\vec{b} = 4\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{c} = \hat{i} + \hat{j} + \hat{k}$ is:

In the following case,find the distance of the given point from the corresponding given plane.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

Find the equation of the set of points which are equidistant from the points $(1, 2, 3)$ and $(3, 2, -1)$.

Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $(-\lambda^2, 1, 1)$,$(1, -\lambda^2, 1)$,and $(1, 1, -\lambda^2)$ also passes through the point $(-1, -1, 1)$. Then $S$ is equal to