3. Given $|y-2| \leqslant 1$, and $x+y=3$. Then the minimum value of $2 x^{2}-$ $12 x+y^{2}+1$ is $\qquad$ .


3. -14 .

It is easy to get $1 \leqslant y \leqslant 3, y=3-x$.
Also, $M=2(x-3)^{2}+y^{2}-17=3 y^{2}-17$, it is clear that when $y=1$, $M$ is the smallest, and the minimum value is -14.