9. If $x-1=\frac{y+1}{2}=\frac{z-2}{3}$, then the minimum value of $x^{2}+y^{2}+z$ is ( ).
(A) 3
(B) $\frac{59}{14}$
(C) $\frac{9}{2}$
(D) 6
9. B.
$$
\begin{array}{l}
\text { Let } x-1=\frac{y+1}{2}=\frac{z-2}{3}=k \text {, then } \\
x=k+1, y=2 k-1, z=3 k+2 . \\
x^{2}+y^{2}+z^{2}=14 k^{2}+10 k+6 \\
=14\left(k+\frac{5}{14}\right)^{2}+\frac{59}{14} .
\end{array}
$$