Q.Find the vector equation of the plane which bisects the segment joining A(2,3,6) and B(4,3,-2) at right angle.

[!TLDR] Perpendicular bisector plane: normal $=\vec{AB}$, point $=$ midpoint. …

$A(2,3,6)$, $B(4,3,-2)$. Midpoint $M=\left(\dfrac{2+4}{2},\dfrac{3+3}{2},\dfrac{6-2}{2}\right)=(3,3,2)$.

Since the plane bisects $AB$ at right angles, its normal is along $\vec{AB}=B-A=(2,0,-8)$, simplified to $(1,0,-4)$, and it passes through $M(3,3,2)$. …

The perpendicular bisector plane of a segment has the segment's own direction vector as normal …

Using $A$ or $B$ as the point instead of the midpoint (only the midpoint lies on the perpendicular-bisector plane); forgetting to simplify $(2,0,-8)$ to $(1,0,-4)$ (not required, but keeps numbers small — using th …