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

If the planes $3x - 2y + 2z + 17 = 0$ and $4x + 3y - kz = 25$ are mutually perpendicular,then $k = $

Let the plane $\pi$ pass through the point $(1,0,1)$ and be perpendicular to the planes $2x+3y-z=2$ and $x-y+2z=1$. Let the equation of the plane passing through the point $(11,7,5)$ and parallel to the plane $\pi$ be $ax+by-z-d=0$. Then,$\frac{a}{b}+\frac{b}{d}=$

$A$ point on the plane determined by the points $A(1,1,-1)$,$B(2,-1,0)$,and $C(-1,0,2)$ among the following is:

Find the vector equation of the plane passing through the points $R(2, 5, -3)$,$S(-2, -3, 5)$,and $T(5, 3, -3)$.