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.)
Let $\mathbf{OP} = <a,b,c> $.
We know that $AB$ and $\mathbf{OP}$ must be perpendicular, so the dot product of $\mathbf{OP}$ and $<2,2,-2>$ is $0$ ($<2,2,-2>$ plays the role of $\mathbf{v}$ in the diagram below)
$$<2,2,-2> \bullet <a,b,c> = 2a+2b-2c = 0$$
$$ \Rightarrow a+b-c=0$$
We also have that for some value of $\lambda$ (let's just call it $t$)
$$<1,2,2> + <2,2,-2>t $$
$$=<1+2t, 2+2t, 2-2t> = <a,b,c>$$
because $P$ lies in the $AB $
Combining the two equations, we get
$$1+2t +2 +2t - 2 +2t = 0 \Rightarrow 6t = -1$$
$$\Rightarrow t = -1/6$$
So $\mathbf{OP} = <2/3, 5/3, 7/3> = \frac13 <2,5,7>$
Best Answer
Let the position vector of the point say $D$ be $\vec{OD}=ai+bj$ then we have $$\vec{AD}=\vec{OD}-\vec{OA}=ai+bj-(i-3j)=(a-1)i+(b+3)j$$ $$\implies |\vec{AD}|=\sqrt{(a-1)^2+(b+3)^2}$$ $$\vec{CD}=\vec{OD}-\vec{OC}=ai+bj-(3i-j)=(a-3)i+(b+1)j$$ $$\implies |\vec{CD}|=\sqrt{(a-3)^2+(b+1)^2}$$ Now, the point $D$ lies on the (extended) line $AC$ then we have $\vec{AD}\parallel \vec{CD}$ then the ratio of corresponding coefficients is constant hence $$\frac{a-1}{a-3}=\frac{b+3}{b+1}$$ $$\implies b=a-4\tag 1$$ Since, the point $D$ divides the line $AC$ in a ratio $-2:3$ or $2:3$ externally. Hence, we have $$\frac{|\vec{AD}|}{|\vec{CD}|}=\frac{2}{3}\implies 9|\vec{AD}|^2-4|\vec{CD}|^2=0$$ setting all the corresponding values, we get $$9((a-1)^2+(b+3)^2)-4((a-3)^2+(b+1)^2)=0$$ setting $b=a-4$ from (1), $$9((a-1)^2+(a-4+3)^2)-4((a-3)^2+(a-4+1)^2)=0$$$$9(a-1)^2-4(a-3)^2=0\implies 5a^2+6a-27=0$$ $$(a+3)(5a-9)=0\implies a=-3, \ \frac{9}{5} $$ Substituting the values of $a$ in (1), we get corresponding values of $b$ as $b=-7$ & $b=-\frac{11}{5}$ respectively.
Hence, the position vector of the point $D$ is $$\color{red}{\vec{OD}=-3i-7j}$$
or $$\color{red}{\vec{OD}=\frac{9}{5}i-\frac{11}{5}j}$$