I have the coordinates of 3 points thrgouh which, a circle should pass . Having the coordinates of the points in 3D, how could I have the coordinates of the center of circumscribed circle ? also : if one of the points has some deviations and causes a circumscribed circle couldn't pass through the 3 points, is there a way to determine the required coordinates of the third point in a way that the circle could be constructed to find the deviation in space ?
Solve the following four equations; here, A, B and C are the 3D-vectors of your three points. M is the center of the circle (what you are looking for), r is its unknown radius. x means "vector product".
 (A-M)² = r²  (B-M)² = r²  (C-M)² = r²  ((C-A)x(C-B))(C-M) = 0
The first three equations say that the distance from the each point to the center must be r. These three conditions will, in 3D, have as solution a straight line of the centers of all spheres that go through points A,B,C.
To find the center of a circle, one must force the solution to the plane where A,B,C lie. There are many ways to accomplish this; the following one seems the easiest one: First, one creates the plane's normal vector: Two vectors in the plane are C-A and C-B, and their cross product is a normal vector. To force C-M (and hence M) also into the plane, the scalar product with normal vector must be 0 (so that they are at right angles, i.e., the normal vector of the plane is orthogonal to C-M).
A=(1,1,1), B=(2,3,1), C=(3,3,2)
Here are the four equations:
 (1-x)² + (1-y)² + (1-z)² = r²  (2-x)² + (3-y)² + (1-z)² = r²  (3-x)² + (3-y)² + (2-z)² = r² (2,2,1) x (1,0,1) = (2,-1,-2), so  2(3-x) - (3-y) - 2(2-z) = 0
With all parentheses expanded:
 1-2x+x² + 1-2y+y² + 1-2z+z² = r²  4-4x+x² + 9-6y+y² + 1-2z+z² = r²  9-6x+x² + 9-6y+y² + 4-4z+z² = r²  6-2x - 3+y -4+2z = 0
To get rid of all the squares in ..., subtract subsequent equations:
- is a linear equation in x,y,z
-3+2x -8+4y = 0
- is also a linear equation in x,y,z
-8+4x -8+4y -3+2z = 0
 is anyway a linear equation in x,y,z
6-2x - 3+y -4+2z = 0
This is now a simple system of three linear equations with three variables, which is easy to solve. The solution yields the coordinates of M:
x = 13/6, y = 5/3, z = 11/6
r can be computed from any of ...:
r = square root of ((1-13/6)² + (1-5/3)² + (1-11/6)²) = square root of 5/2
And as joojaa pointed out above, you can always find a circle through 3 points (unless they are on a straight line, which will make the equations above unsolvable).