Let the three coordinates be (A(x_1,y_1,z_1)), (B(x_2,y_2,z_2)), (C(x_3,y_3,z_3))
Form two direction vectors in the plane:
(vec{AB}=(x_2-x_1, y_2-y_1, z_2-z_1))
(vec{AC}=(x_3-x_1, y_3-y_1, z_3-z_1))
Compute the normal vector with the cross product:
(vec{n}=vec{AB}timesvec{AC}=(a,b,c))
Write the plane equation using point (A):
(a(x-x_1)+b(y-y_1)+c(z-z_1)=0)
Expand to standard form:
(ax+by+cz+d=0)
Compute (d) using one point:
(d=-(ax_1+by_1+cz_1))
Final plane equation:
(ax+by+cz+d=0)
If the cross product is ((0,0,0)), the three points are collinear and no unique plane exists