Let $$\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}, \vec{b}=-\hat{i}+\hat{k}, \vec{c}=\beta \hat{j}-\hat{k}$$, where $\alpha$ and $\beta$ are integers and $\alpha \beta=-6$. Let the values of the ordered pair $(\alpha, \beta)$, for which the area of the parallelogram of diagonals $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is $\frac{\sqrt{21}}{2}$, be $$\left(\alpha_1, \beta_1\right)$$ and $$\left(\alpha_2, \beta_2\right)$$. Then $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ is equal to