Given points $A(3, 2, -1)$ and $B(1, 4, 3)$,find the equation of the plane that bisects the segment $AB$ perpendicularly.

The distance of the plane $\vec{r} = (\hat{i} - \hat{j}) + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k})$ from 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 $P$?

The line drawn from $(4, -1, 2)$ to the point $(-3, 2, 3)$ meets a plane at right angles at the point $(-10, 5, 4)$. Find the equation of the plane.

$A$ plane which bisects the angle between the two given planes $2x - y + 2z - 4 = 0$ and $x + 2y + 2z - 2 = 0$,passes through the point

Let $\pi$ be the plane passing through the point $(3,-3,1)$ and perpendicular to the line joining the points $(3,4,-1)$ and $(2,-1,5)$. If the equation of the plane containing the points $(3,4,-1),(-1,2,5)$ and perpendicular to the plane $\pi$ is $ax+y+cz-d=0$,then $3(a+c)=$