There are three point coordinates $A(x_1,y_1,z_1),B(x_2,y_2,z_2),C(x_3,y_3,z_3)$, where $z_1=z_2=z_3$ Now,
$(x-x_1)^2+(y-y_1)^2+(z-z_1)^2=d_1^2$ this equation can be written for my system as
$(x-x_1)^2+(y-y_1)^2+h^2=c_1 \cdot h^2$(where $c_1$ is a constant and $h$ is a unknown variable mainly vertical distance) so for three point i get three equation like
$(x-x_1)^2+(y-y_1)^2+h^2=(c_1 \cdot h)^2$
$(x-x_2)^2+(y-y_2)^2+h^2=(c_2 \cdot h)^2$
$(x-x_3)^2+(y-y_3)^2+h^2=(c_3 \cdot h)^2$
now i have three equation and three unknown variables $(x,y,h)$ to find out the three unknown by using numerical method i used least square estimation method ( i don't know any other numerical method for these kind of equations) , for using least square estimation i used $2 \cdot M^T \cdot M \cdot X=M^T \cdot D$ so i need another equation for point D$(x_4,y_4,z_4)$ also $z_4=z_1=z_2=z_3$ and equation $(x-x_4)^2+(y-y_4)^2+h^2=(c_4 \cdot h)^2$
in least square equation where
$$M=\begin{pmatrix}(x_4-x_1)&(y_4-y_1)&(z_4-z_1)\\(x_4-x_2)&(y_4-y_2)&(z_4-z_2)\\(x_4-x_3)&(y_4-y_3)&(z_4-z_3)\end{pmatrix}$$
and $$D=\begin{pmatrix}(d_1^2-d_4^2)-(x_1^2-x_4^2)-(y_1^2-y_4^2)-(z_1^2-z_4^2)\\(d_2^2-d_4^2)-(x_2^2-x_4^2)-(y_2^2-y_4^2)-(z_2^2-z_4^2)\\(d_3^2-d_4^2)-(x_3^2-x_4^2)-(y_3^2-y_4^2)-(z_3^2-z_4^2)\end{pmatrix}$$
when i substitute my equation in M and D these are look like
$$M=\begin{pmatrix}(x_4-x_1)&(y_4-y_1)&(h+z-h-z)\\(x_4-x_2)&(y_4-y_2)&(h+z-h-z)\\(x_4-x_3)&(y_4-y_3)&(h+z-h-z)\end{pmatrix}$$
and
$$D=\begin{pmatrix}((c_1 \cdot h)^2-(C_4 \cdot h)^2)-(x_1^2-x_4^2)-(y_1^2-y_4^2)-((h+z)^2-(h+z)^2))\\((c_2 \cdot h)^2-(C_4 \cdot h)^2)-(x_1^2-x_4^2)-(y_1^2-y_4^2)-((h+z)^2-(h+z)^2))\\((c_3 \cdot h)^2)-(C_4 \cdot h)^2)-(x_1^2-x_4^2)-(y_1^2-y_4^2)-((h+z)^2-(h+z)^2))\end{pmatrix}$$
at this moment h is eliminate from M and h exist in D so how can i find the value of x,y and h. Is there any other method to solve this equation i prefer by using only three and must be numerical method no subtraction or multiplication method because in that case error is large.
Let us consider your three equations (written as A=0, B=0, C=0). If you substract A from B and A form C expanding the squares, you vill notice that terms x^2 and y^2 disappear and that you just have two linear equations for two unknownx x and y (considereding h as a constant). So, you have x and y as analytical expressions of h. Replace these expressions now in equation A and you have a single equation of h (which is suppose is a quadratic).
If you want to use a least square estimation method, just keep your three equations and three unknowns. At solution, the sum of squares must be zero since you have as many equations and unknowns. Do not add any extra.