There are these three quadratic equations:
$(x - x_1)^2 + (y - y_1)^2 + P_1 \cdot h^2 = 0$ ……..(1)
$(x - x_2)^2 + (y - y_2)^2 + P_2 \cdot h^2 = 0$ ……..(2)
$(x - x_3)^2 + (y - y_3)^2 + P_3 \cdot h^2 = 0$ ……..(3)
where we know the values of $x_1, x_2, x_3, y_1, y_2, y_3, P_1, P_2, P_3$.
How can I find $(x,y,h)$?
Subtract (1) from (2) to get a linear equation in $x,y$
Subtract (1) from (3) to get another
From these solve for $x$ and $y$.
Substitute back in (1) to get a quadratic in $h^2$.
Solve to get $h^2$
Note till the very end, you can call $z=h^2$ and finally solve for $h$
Added in response to comments
Here is a worked out example (the following is computer generated so excuse the poor formatting).
Let the equations be $$\left(y-6\right)^2+\left(x-9\right)^2-{{5\,h^2}\over{64}} \tag 1 $$ $$ \left(y-9\right)^2+\left(x-19\right)^2-{{61\,h^2}\over{32}} \tag2 $$ $$ \left( y-7\right)^2+\left(x-1\right)^2-{{25\,h^2}\over{32}} \tag 3$$
Subtract (1) from (2) and also (1) from (3) and simplify to get $$ -384\,y-1280\,x-117\,h^2+20800 =0 \\ -128\,y+1024\,x-45\,h^2- 4288 =0 $$
Solving the two equations, letting $z=h^2$ we get $$x={{9\,z+16832}\over{2176}} , ~~~y=-{{693\,z-61760}\over{2176}} \tag 4 $$ Now substitute in (1) to get $$240165\,z^2-33961600\,z+1189826560=0$$ Solving $$ z={{3718208}\over{48033}} , z=64 $$ which gives $$ h = \pm {{8\,\sqrt{58097}}\over{9\,\sqrt{593}}}, h=\pm 8$$ Substituting in (4) we get $$x={{42992}\over{5337}} , y={{19904}\over{5337}} , h=- {{8\,\sqrt{58097}}\over{9\,\sqrt{593}}} \\ x={{42992 }\over{5337}} , y={{19904}\over{5337}} h={{8\,\sqrt{58097}}\over{9 \,\sqrt{593}}} \\x=8 , y=8 , h=8 \\ x= 8 , y=8 , h=-8 $$