If the point $(2, \alpha, \beta)$ lies on the plane which passes through the points $(3, 4, 2)$ and $(7, 0, 6)$ and is perpendicular to the plane $2x - 5y = 15$,then $2\alpha - 3\beta$ is equal to:

$A$ vector $n$ of magnitude $8$ units is inclined to $x$-axis at $45^\circ$,$y$-axis at $60^\circ$ and an acute angle with $z$-axis. If a plane passes through a point $(\sqrt{2}, -1, 1)$ and is normal to $n$,then its equation in vector form is

Find the Cartesian equation of the following plane: $\vec{r} \cdot (2\hat{i} + 3\hat{j} - 4\hat{k}) = 1$

If $P$ is the point $(2, 6, 3)$,then the equation of the plane passing through $P$ and perpendicular to $OP$,where $O$ is the origin,is:

Let the acute angle bisector of the two planes $x-2y-2z+1=0$ and $2x-3y-6z+1=0$ be the plane $P$. Then which of the following points lies on $P$?

If the planes $2x + 3y + 4z + 7 = 0$ and $4x + ky + 8z + 1 = 0$ are parallel,then the equation of the plane passing through the point $(k, k, k)$ and having the direction ratios of its normal as $(k-1, k, k+1)$ is