OP indicates they already know the location of the point where the three spheres intersect, projected to the plane through the three sphere centers; only the distance $w$ from the plane to the intersection is unknown.
Simply put, the distance $w$ is the difference between the sphere radius and the distance from the projected point to the center of that sphere, using any of the three spheres.
Let $(x_p, y_p, z_p)$ be the point on the plane, $r_1$ be the radius of one of the spheres, and $(x_1, y_1, z_1)$ be the center of that sphere. Then,
$$w = \sqrt{ r_1^2 - (x_p - x_1)^2 - (y_p - y_1)^2 - (z_p - z_1)^2 }$$
Here is the full trilateration procedure:
Let's say you have three spheres with radii $r_1$, $r_2$, and $r_3$, centered at $\vec{c}_1 = ( x_1 , y_1 , z_1 )$, $\vec{c}_2 = ( x_2 , y_2 , z_2 )$, and $\vec{c}_3 = ( x_3 , y_3 , z_3 )$.
To simplify the problem, we rotate the spheres so that their centers lie on the $x y$ plane, and translate so that the center of the first sphere is at origin $(0, 0, 0)$, the center of the second sphere is on the positive $x$ axis at $(u_2, 0, 0)$ with $u_2 \gt 0$, and the center of the third sphere is at $(u_3, v_3, 0)$, with $v_3 \gt 0$.
To make it clearer which orientation is involved, I'll use $x$, $y$, and $z$ for the original 3D coordinates, and $u$, $v$, $w$ for the rotated 3D coordinates.
Note that if $u_2 = 0$, the first and second sphere are centered on the same point, and there is either no solution, or an infinite number of solutions (a circle instead of one or two points).
Similarly, if $u_3 = v_3 = 0$, the first and third spheres are centered on the same point. If $v_3 = 0$ (but $u_3 \ne 0$), the three sphere centers are collinear, and again there is either no solution, or the solutions form a circle.
In the sane, non-degenerate cases, $u_2 \gt 0$ and $v_3 \gt 0$, and that is what we assume here.
We can obtain the unit axis vectors for the rotated coordinate system, $\hat{u}$, $\hat{v}$, and $\hat{w}$, from the vectors between the sphere centers, using the Gram-Schmidt process:
$$\begin{aligned}
\vec{d}_{12} &= \vec{c}_2 - \vec{c}_1 \\
\vec{d}_{13} &= \vec{c}_3 - \vec{c}_1 \\
L_{12} &= \sqrt{\left\lVert \vec{d}_{12} \right\rVert} \\
L_{13} &= \sqrt{\left\lVert \vec{d}_{13} \right\rVert} \\
\hat{u} &= \frac{\vec{d}_{12}}{L_{12}} \\
\vec{v} &= \vec{d}_{13} - \hat{u} ( \hat{u} \cdot \vec{d}_{13} ) \\
\hat{v} &= \frac{\vec{v}}{\left\lVert \vec{v} \right\rVert} \\
\hat{w} &= \hat{u} \times \hat{v} \\
\end{aligned}$$
Note that this rotated coordinate system is right-handed. (Its handedness doesn't actually matter; I only bring it up because if your original coordinate system is left-handed, then this new coordinate system is a generalized rotation that involves mirroring one of the axes, instead of a pure rotation.)
Note that in the original coordinate system,
$$\begin{aligned}
\vec{c}_2 &= \vec{c}_1 + u_2 \hat{u} \\
\vec{c}_3 &= \vec{c}_1 + u_3 \hat{u} + v_3 \hat{v} \\
\end{aligned}$$
Because the Gram-Schmidt process makes sure that $\hat{u}\cdot\hat{v} = 0$, we directly obtain the sphere center coordinates in the rotated coordinate system from the above:
$$\left\lbrace \begin{aligned}
u_2 &= L_{12} \\
u_3 &= \hat{u} \cdot \vec{d}_{13} \\
v_3 &= \hat{v} \cdot \vec{d}_{13} \\
\end{aligned} \right.$$
Because the three spheres are on the $u v$ plane, the two points where all three spheres intersect are $(u, v, \pm w)$, and
$$\left\lbrace \begin{aligned}
u^2 + v^2 + w^2 &= r_1^2 \\
(u - u_2)^2 + v^2 + w^2 &= r_2^2 \\
(u - u_3)^2 + (v - v_3)^2 + w^2 &= r_3^2 \\
\end{aligned} \right.$$
The first of the three simultaneous equations restricts the intersection point to the shell of the first sphere, the second to the shell of the second sphere, and the third to the shell of the third sphere. The solution to this system of equations is
$$\left\lbrace \begin{aligned}
u &= \frac{r_1^2 - r_2^2 + u_2^2}{2 u_2} \\
v &= \frac{r_1^2 - r_3^2 + L_{13}^2 - 2 u_3 u}{2 v_3} \\
w &= \sqrt{r_1^2 - u^2 - v^2 } \\
\end{aligned} \right.$$
and the three spheres intersect at
$$\vec{c}_1 + u \hat{u} + v \hat{v} \pm w \hat{w}$$
in the original coordinate system.
The picture above represents my triangulation of 3 intersecting spheres in 2D space. This is the known. What I'm trying to do is find is now find the height in which all 3 intersect. The below picture is my best attempt at re-creating this.
