If the equation of the plane passing through the point $A(-2, 1, 3)$ and perpendicular to the vector $3 \hat{i} + \hat{j} + 5 \hat{k}$ is $ax + by + cz + d = 0$,then $\frac{a + b}{c + d} = $

$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$. 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

The equation of the plane which bisects the angle between the planes $3x - 6y + 2z + 5 = 0$ and $4x - 12y + 3z - 3 = 0$ which contains the origin is

$A$ plane passes through the point $(1, -2, 1)$ and is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$. Find the distance of the plane from the point $(1, 2, 2)$.

Let $6x - 3y + 2z - 6 = 0$ be the given plane. If $a, b, c$ are the intercepts made by the plane on $X, Y, Z$-axes respectively; $l, m, n$ are the direction cosines of a normal drawn to the plane and $p$ is the perpendicular distance from the origin to the plane, then $|al + bm + cn|=$