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 $(x_0, y_0, z_0)$?

Numerically attempted a solution. Computed the circle center, which is also circum-center of the triangle formed by joining the three assumed points.. in the program below. The sides are found, the circumradius R is calculated using trigonometric relations.
R is then equated to each of the three center to vertex distances to get the circle center, conventional way by solving three equations for the center.
Did not have luck so far finding the center coordinates analytically. Any help for generalizing the above in Mathematica or in other ways would be much appreciated.
x1=3;y1=5.;z1=3.2;x2=3;y2=5;z2=.8;x3=1.2;y3=.7;z3=5.2;a=Sqrt[(x1-x2)^2+(y1-y2)^2+(z1-z2)^2];b=Sqrt[(x2-x3)^2+(y2-y3)^2+(z2-z3)^2];c=Sqrt[(x3-x1)^2+(y3-y1)^2+(z3-z1)^2];s=(a+b+c)/2;Area=Sqrt[s(s-a)(s-b)(s-c)];R=a b c/4/Area;eq={(xc-x1)^2+(yc-y1)^2+(zc-z1)^2==R^2,(xc-x2)^2+(yc-y2)^2+(zc-z2)^2==R^2,(xc-x3)^2+(yc-y3)^2+(zc-z3)^2==R^2};NSolve[eq,{xc,yc,zc}]
                                                                                                                                                                                                                                                                                                                                  
Out[41]= {{xc->1.73553,yc->1.97931,zc->2.},{xc->1.73553,yc->1.97931,zc->2.}}

