The equation of the plane passing through $(2, 3, 4)$ and parallel to the plane $5x - 6y + 7z = 3$ is:

$P$ and $Q$ are points on the straight line passing through the point $A(3 \hat{i}+\hat{j}-\hat{k})$ and parallel to the vector $\vec{v} = 2 \hat{i}-\hat{j}+2 \hat{k}$. If $AP = AQ = 3$,then the vector equation of the plane $OPQ$ is:

Prove that if a plane has the intercepts $a, b, c$ and is at a distance of $p$ units from the origin,then $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}}$.

Let $P$ be the plane passing through the point $(1, 2, 3)$ and the line of intersection of the planes $\vec{r} \cdot (\hat{i} + \hat{j} + 4\hat{k}) = 16$ and $\vec{r} \cdot (-\hat{i} + \hat{j} + \hat{k}) = 6$. Then which of the following points does $NOT$ lie on $P$?

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:

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 $|al + bm + cn|=$