Let the points on the plane $P$ be equidistant from the points $A(-4, 2, 1)$ and $B(2, -2, 3)$. Then the acute angle between the plane $P$ and the plane $2x + y + 3z = 1$ is

$A$ 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

Let $P$ be the plane passing through the points $(5,3,0), (13,3,-2)$ and $(1,6,2)$. For $\alpha \in N$,if the distances of the points $A(3,4,\alpha)$ and $B(2,\alpha,a)$ from the plane $P$ are $2$ and $3$ respectively,then the positive value of $a$ is:

The mirror image of the point $A(1, 2, 3)$ in a plane is $B\left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$. Which of the following points lies on this plane?

Find the equation of the plane that passes through the three points $(1,1,0), (1,2,1),$ and $(-2,2,-1)$.

$(1, -2, 1)$ is a point on a plane $\pi$ and $\pi$ is parallel to the plane $x-y-z=0$. If the equation of $\pi$ is $ax+by+cz-2=0$,then $b-2c=$