The perpendicular distance of the origin from the plane $x-3y+4z-6=0$ is

The vector equation of the plane $r = (2 \hat{i} + \hat{k}) + \lambda(\hat{i}) + \mu(\hat{i} + 2 \hat{j} - 3 \hat{k})$ in scalar product form is $r \cdot (3 \hat{i} + 2 \hat{k}) = \alpha$,then $\alpha = \dots$

The equation of the planes parallel to the plane $x - 2y + 2z - 3 = 0$ which are at a unit distance from the point $(1, 2, 3)$ is $ax + by + cz + d = 0$. If $(b - d) = K(c - a)$,then the positive value of $K$ is

If $l, m, n$ are the direction cosines of a normal to the plane passing through the points $(0, 1, 2)$, $(3, 0, 2)$, and $(4, 5, 0)$, then $|l| + |m| + |n| = $

Find the distance of the point $(0, 0, 0)$ from the plane $3x - 4y + 12z = 3$. (in $/13$)

The distance from a point $(1, 1, 1)$ to a variable plane $\pi$ is $12$ units and the points of intersections of the plane $\pi$ and $X, Y, Z$-axes are $A, B, C$ respectively. If the point of intersection of the planes through the points $A, B, C$ and parallel to the coordinate planes is $P$,then the equation of the locus of $P$ is