9. If real numbers $x, y$ satisfy $3 x+2 y-1 \geqslant 0$, then the minimum value of $u=x^{2}+y^{2}+6 x-2 y$ is $\qquad$ .
9. $-\frac{66}{13}$.
$u=(x+3)^{2}+(y-1)^{2}-10$. The minimum distance from a point in the half-plane $3 x+2 y-1$ $\geqslant 0$ to the fixed point $(-3,1)$ is $\frac{|-9+2-1|}{\sqrt{13}}=\frac{8}{\sqrt{13}}$. Therefore,
$$
u_{\text {min }}=\left(\frac{8}{\sqrt{13}}\right)^{2}-10=-\frac{66}{13} .
$$