Area of parallelogram whose diagonals are $$\vec{a}+\vec{b}$$ and $$\vec{b}+\vec{c}$$ is

$$\begin{aligned} & =\frac{1}{2}|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})| \\ & =\frac{1}{2}|\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}| \\ & =\frac{1}{2}|-2 \beta \hat{i}-2 \hat{j}+(\alpha+\beta) \hat{k}| \end{aligned}$$

$$=\frac{1}{2} \sqrt{4 \beta^2+4+(\alpha+\beta)^2}$$

Which is given $$\frac{\sqrt{21}}{2}$$

$$\begin{array}{ll} \therefore & 4 \beta^2+4+(\alpha+\beta)^2=21 \\ \Rightarrow & (\alpha+\beta)^2+4 \beta^2=17 \\ \Rightarrow & \alpha^2+5 \beta^2+2 \alpha \beta=17 \\ \Rightarrow & \alpha^2+5 \beta^2=29 \\ \therefore & (\alpha, \beta) \in\{(3,2),(-3,-2),(-3,2),(3,-2)\} \\ \because & \alpha \beta=-6 \\ \therefore & (\alpha, \beta) \in\{(-3,2),(3,-2)\} \\ \therefore & \alpha_1^2+\beta_1^2-\alpha_2 \beta_2 \\ = & 9+4-(-6)=19 \end{array}$$