(14) 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 $\qquad$
$14-\frac{66}{13}$ Hint: From $u=x^{2}+y^{2}+6 x-2 y=(x+3)^{2}+(y-1)^{2}-10$, 
we know that the geometric meaning of $u$ is the square of the distance from any point in the feasible region to the point $(-3,1)$ minus 10.