If the vertices $A, B, C$ of a triangle $ABC$ are $(1, 2, 3), (-1, 0, 0), (0, 1, 2)$ respectively, then find $\angle ABC. \angle ABC$ is the angle between the vectors $\overline {BA}$ and $\overline {BC}$.
$\begin{array}{1 1}(A) \cos^{-1}\big(\large\frac{10}{\sqrt{102}}\big) \\ (B) \cos^{-1}\big(\large\frac{-10}{\sqrt{102}}\big) \\(C) \cos^{-1}\big(\large\frac{10}{102}\big) \\ (D) \cos^{-1}\big(\large\frac{-10}{102}\big)\end{array}$