3D coordinates of circle center given three point on the circle. Given the three coordinates $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, $(x_3, y_3, z_3)$ defining a circle in 3D space, how to find the coordinates of the center of the circle $(x_0, y_0, z_0)$?

I think using a projection into 2D might be the easiest way to actually calculate. If the points are $A, B, C$ then find 
$$\begin{align}
\mathbf{u_1} & = B-A \\
\mathbf{w_1} &= (C-A) \times \mathbf{u_1} \\
\mathbf{u} & = \mathbf{u_1} / | \mathbf{u_1} | \\
\mathbf{w} & = \mathbf{w_1} / | \mathbf{w_1} | \\
\mathbf{v} & = \mathbf{w} \times \mathbf{u} \\
\end{align}$$
This gives three orthogonal unit vectors with $\mathbf{u}$ and $\mathbf{v}$ spanning the plane. 
Get 2D coordinated by taking the dot products of $(B-A)$ and $(C-A)$ with $\mathbf{u}$ and $\mathbf{v}$. Let 
$$\begin{align}
b & = (b_x,0) = ( (B-A) \cdot \mathbf{u} , 0 ) \\
c &= (c_x,c_y) = ( (C-A) \cdot \mathbf{u}, (C-A)\cdot \mathbf{v} ) \\
\end{align}$$
We know the center must lie on the line $x= b_x/2$. Let this point be $(b_x/2,h)$. The distance from c must be the same as the distance from the origin
$$(c_x-b_x/2)^2 + (c_y - h)^2 = (b_x/2)^2 + h^2$$
So 
$$h = \frac{(c_x-b_x/2)^2 + c_y^2 - (b_x/2)^2}{ 2 c_y } $$
The actual center can then be recovered by taking $A + (b_x/2)\mathbf{u} + h \mathbf{v}$.
This is a nice explicit calculation. 
