3. If real numbers $x, y$ satisfy $4 x^{2}-4 x y+2 y^{2}=1$, then the sum of the maximum and minimum values of $3 x^{2}+x y+y^{2}$ is $\qquad$
Given $3 x^{2}+x y+y^{2}=k=k\left(4 x^{2}-4 x y+2 y^{2}\right)$, then $(4 k-3) x^{2}-(4 k+1) x y+(2 k-1) y^{2}=0$. If $k=\frac{3}{4}, y^{2}=8 x y$, at this time $(x, y)=\left( \pm \frac{1}{2}, 0\right)$ or $\left( \pm \frac{1}{10}, \pm \frac{4}{5}\right)$; if $k \neq \frac{3}{4}$, by $y \neq 0 \Rightarrow(4 k-3)\left(\frac{x}{y}\right)^{2}-(4 k+1) \cdot \frac{x}{y}+(2 k-1)=0$, then $\Delta=(4 k+1)^{2}-4(4 k-3)(2 k-1) \geqslant 0 \Rightarrow 16 k^{2}-48 k+11 \leqslant 0 \Rightarrow \frac{1}{4} \leqslant k \leqslant \frac{11}{4}$. Therefore, the maximum and minimum values of $3 x^{2}+x y+y^{2}$ are $\frac{11}{4}$ and $\frac{1}{4}$, respectively, and their sum is 3.