$A$ plane $\pi$ passing through the point $(1,1,1)$ is perpendicular to the line joining the points $(6,3,2)$ and $(1,-4,-9)$. If $ax+by+cz-23=0$ is the equation of the plane $\pi$, then $a+b-c=$

If $(2, -1, 3)$ is the foot of the perpendicular drawn from the origin $(0, 0, 0)$ to a plane,then the equation of that plane is:

If the distance between the planes $2x + y + z + 1 = 0$ and $2x + y + z + \alpha = 0$ is $3$ units,then the product of all possible values of $\alpha$ is

$A$ plane $\pi$ makes intercepts $3$ and $4$ respectively on $Z$-axis and $X$-axis. If $\pi$ is parallel to $Y$-axis, then its equation is:

The plane $ax + by = 0$ is rotated about its line of intersection with the plane $z = 0$ through an angle $\alpha$. Prove that the equation of the plane in its new position is $ax + by \pm (\sqrt{a^{2} + b^{2}} \tan \alpha) z = 0$.

$A$ plane passes through $(2,1,2)$ and $(1,2,1)$ and is parallel to the line $2x = 3y$ and $z = 1$. Then the plane also passes through which of the following points?