If you always want the $(x-x_1,y-y_1)$ vector to be rotated $90°$ counter-clockwise against the $(x_2-x_1,y_2-y_1)$ one, then you have
\begin{align*}
n(x-x_1) &= m(y_1-y_2) \\
n(y-y_1) &= m(x_2-x_1)
\end{align*}
So the left $y$ difference is the right $x$ difference, and the left $x$ difference is the negative of the right $y$ difference. This amounts to the $90°$ rotation I mentioned.
Solve the above and you find
\begin{align*}
x &= \frac{m(y_1-y_2)}n+x_1 \\
y &= \frac{m(x_2-x_1)}n+y_1
\end{align*}
which corresponds to one of your four solutions.
Let $\overrightarrow{B} . \overrightarrow{A}=0$
I call $x_3$ the $x$ coordinate of vertex $v_3$, for convenience purpose.
That is $(x_3-x_2)(x_2-x_1)+(y_3-y_2)(y_2-y_1)+(z_3-z_2)(z_2-z_1)=0$
Given that you know $x_1,x_2,x_3,y_1, y_2, z_1,z_2$ it gives you a relationship between $y_3$ and $z_3$ (or, easier to use - see afterwards, a relationship between $(y_3-y_2)$ and $(z_3-z_2)$.
Then you know that the lenght of $\overrightarrow{B}$ is $0.99$
Then you have $(x_3-x_2)^2+(y_3-y_2)^2+(z_3-z_2)^2=0.99^2$
Granted, this gives you a potential of two values for $y_3$, but you should be able to rule out one of the two if needed.
You could have used the scalar product between $\overrightarrow{B}$ and $\overrightarrow{C}$, that would have given you a non-quadratic equation, but that I feel more complex to compute.
Best Answer
Let's continue with your method. Drop a perpendicular from $A$ of the line $BC$ and call the point $M$. Using similar triangles (which does help), we have:
$$\frac{BM}{AM} = \frac{AM}{CM}$$ $$\frac{3}{3b} = \frac{3b}{12}$$ $$9b^2=36 ,b=2$$
which explains where the formula $h^2=pq$ comes from.
Using similar triangles again, we have:
$$\frac{AM}{CM} = \frac{QP}{PC}$$ $$\frac{3 \cdot 2}{ 12} = \frac{QP}{9}$$
and you can surely continue from here.