14. Given 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. $-\frac{66}{13}$.

From $u=x^{2}+y^{2}+6 x-2 y$
$$
=(x+3)^{2}+(y-1)^{2}-10 \text {, }
$$

we know that the geometric meaning of the function $u$ is the square of the distance from any point in the feasible region to the point $(-3,1)$, minus 10.