Find the equation $ax + by + cz = d$ of the plane which has equal distance to the points $A(1, 2, 3)$ and $B(4, 5, 6)$ I was just wondering if anyone has any suggestions as to how to compute this equation?

Find the equation $ax + by  + cz = d$ of the plane for which every point has equal
  distance to the points $A(1, 2, 3)$ and $B(4, 5, 6)$

 A: The vector$\overrightarrow{AB}=(3,3,3)$, hence the vector $(1,1,1)$,  is normal to the perpendicular bisector plane. Thus its equation is $\;x+y+z=d$.
Furthermore, it passes through the midpoint of $(AB)$ which has coordinates $\;\Bigl(\dfrac52,\dfrac72,\dfrac92\Bigr)$. Finally the equation is:
$$x+y+z=\frac{21}2.$$
A: Then AB in orthogonal to the plane, so 
$a=4-1, b=5-2, c = 6-3$ and the middle of the AB must be on the plane (this will give you d)
A: HINT:
The Plane must be on the middle point of A and B.


*

*Find the middle point C of A and B

*Use the fact that the normal vector of the plane is $\vec{AB}$ 


Not: A normal vector of a plane ($ax + by  + cz = d$) is $\vec{N}=(a,b,c)$


*Satify Point C in the plane equation ($ax + by  + cz = d$).

A: Let $I$ be the medium of the segment $[AB].$ Then $I(5/2; 7/2; 9/2).$ Let $M(x;y;z)$ be a point from this plan than we have the following inner product is zero i.e $AB.IM=0.$ So, $3(x-5/2)+3(y-7/2)+3(z-9/2)=0$ which implies that 
$$x+y+Z=21/2.$$ Thus, $a=b=c=1$ and $d=21/2.$
