It is likely more helpful to set out the problem using vector forms for the relevant lines. The vector $ \ \vec{c} \ = \ \vec{AB} \ $ is $ \ \vec{c} \ = \ \vec{b} \ - \ \vec{a} \ $ (since $ \ \vec{a} \ + \ \vec{c} \ = \ \vec{b} \ $ ) . It appears that you are working in $ \ \mathbb{R}^2 \ $ , so we have
$$ \vec{a} \ = \ \langle x_1 \ , \ y_1 \rangle \ \ , \ \ \vec{b} \ = \ \langle x_2 \ , \ y_2 \rangle \ \ , \ \ \vec{c} \ = \ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \ . $$
It is convenient to use $ \ \vec{a} \ $ as the "initial" vector with a parameterization that places $ \ t = 0 \ $ there (point A) and places $ \ t = 1 \ $ at $ \ \vec{b} \ $ (point B) , with the direction given by $ \ \vec{c} \ $ . The vector $ \ \vec{c} \ $ then lies along the line AB having parametric equation
$$ \ \vec{r} \ = \ \vec{a} \ + \ t \ \vec{c} \ = \ \langle x_1 \ , \ y_1 \rangle \ + \ t \ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \ . $$
The midpoint is then located at $ \ t = \frac{1}{2} \ $ , said point being at the head of position vector $ \ \langle \frac{x_1 \ + \ x_2}{2} \ , $ $ \ \frac{y_1 \ + \ y_2}{2} \rangle \ $ .
The perpendicular bisector of $ \ \overline{AB} \ $ can be defined by a vector which has a zero "dot product" (scalar product) with $ \ \vec{c} \ $ . If we say that this line has a slope $ \ m \ $ , we can write its direction as $ \ \langle 1 \ , \ m \rangle \ $ , and so
$$ \langle x_2 - x_1 \ , \ y_2 - y_1 \rangle \ \cdot \ \langle 1 \ , \ m \rangle \ \ = \ \ (x_2 - x_1 ) \ + \ m \ (y_2 - y_1) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ m \ \ = \ \ - \frac{x_2 - x_1}{y_2 - y_1} \ \ . $$
(Note that this is just what we expect from the "product of slopes of perpendicular lines" theorem, since the direction $ \ \vec{c} \ $ has slope $ \ \frac{y_2 - y_1}{x_2 - x_1} \ $ . )
Since $ \ y_2 - y_1 \ \neq \ 0 \ $ , we can then also write the direction for the perpendicular bisector as $ \ \langle y_2 - y_1 \ , \ x_1 - x_2 \rangle \ $ . (It is also acceptable to attach the "minus sign" to the first component instead -- thus, $ \ \langle y_1 - y_2 \ , \ x_2 - x_1 \rangle \ $ -- which is a vector pointing in the opposite direction on the perpendicular bisector line.
Finally, we can establish a parametrization of the perpendicular bisector with the midpoint of $ \ \overline{AB} \ $ at $ \ \tau = 0 \ $ , making it the "initial" vector for the bisector. The vector form for this line may then be written as
$$ \vec{R} \ = \ \langle \frac{x_1 \ + \ x_2}{2} \ , \ \frac{y_1 \ + \ y_2}{2} \rangle \ + \ \tau \ \langle y_2 - y_1 \ , \ x_1 - x_2 \rangle \ \ . $$
This method can be extended to $ \ \mathbb{R}^3 \ $ , except that the perpendicular bisector is ambiguous to the extent that it lies in a plane for which $ \ \vec{AB} \ $ is its normal. To pin things down, we would need to specify a point in that plane that the bisector passes through. (This is sometimes given as a textbook or exam problem.)
Best Answer
The normal is of the form $<x,y,z>=t<3,2,-2>$ So $x=3t,y=2t,z=-2t$ Substitute in the plane equation to find t to be $1/17$ Plug this t value in my equation above to find the point on te plane