The two points $A=(x_{0},y_{0},z_{0})$ and $B=(x_{1},y_{1},z_{1})$ are given. I want to find the coordiantes of the point $C=(x,y,z) $. The line segments $AC$ and $BC$ make equal angle $\alpha$ with the horizontal plane through $C$. The angle $\alpha=\arctan(m)$ which is known.
The question is how to find the location of $C=(x,y,z)$ in terms of $x_{0},y_{0},z_{0},x_{1},y_{1},z_{1}$ and $m$?
First, let us simplify and assume $z = z_0$. Then we would have
$$\frac{y-y_0}{x-x_0} = m \quad \mbox{or} \quad y = m(x-x_0)+y_0$$
It remains to determine $x$
We know that $AC \cdot BC = |AC||BC| \cos(2\alpha)$ so this will give us a quadratic in $x$. Solving this we will get two possible points for $C$.
Now the complete solution will be the circle that has the two points already found as endpoints of a diameter.
I hope you'll excuse me for not working out all the details as the quadratic equation you get in $x$ appears to be quiet messy to work with.