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

<p>Area of parallelogram whose diagonals are $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is</p> <p>$$\begin{aligned} & =\frac{1}{2}|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})| \\ & =\frac{1}{2}|\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}| \\ & =\frac{1}{2}|-2 \beta \hat{i}-2 \hat{j}+(\alpha+\beta) \hat{k}| \end{aligned}$$</p> <p>$=\frac{1}{2} \sqrt{4 \beta^2+4+(\alpha+\beta)^2}$</p> <p>Which is given $\frac{\sqrt{21}}{2}$</p> <p>$$\begin{array}{ll} \therefore & 4 \beta^2+4+(\alpha+\beta)^2=21 \\ \Rightarrow & (\alpha+\beta)^2+4 \beta^2=17 \\ \Rightarrow & \alpha^2+5 \beta^2+2 \alpha \beta=17 \\ \Rightarrow & \alpha^2+5 \beta^2=29 \\ \therefore & (\alpha, \beta) \in\{(3,2),(-3,-2),(-3,2),(3,-2)\} \\ \because & \alpha \beta=-6 \\ \therefore & (\alpha, \beta) \in\{(-3,2),(3,-2)\} \\ \therefore & \alpha_1^2+\beta_1^2-\alpha_2 \beta_2 \\ = & 9+4-(-6)=19 \end{array}$$</p>