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

Let three vectors ,$$\overrightarrow{\mathrm{a}}=\alpha \hat{i}+4 \hat{j}+2 \hat{k}, \overrightarrow{\mathrm{b}}=5 \hat{i}+3 \hat{j}+4 \hat{k}, \overrightarrow{\mathrm{c}}=x \hat{i}+y \hat{j}+z \hat{k}$$ form a triangle such that $$\vec{c}=\vec{a}-\vec{b}$$ and the area of the triangle is $$5 \sqrt{6}$$. If $$\alpha$$ is a positive real number, then $$|\vec{c}|^2$$ is equal to:

Let $$\overrightarrow{O A}=2 \vec{a}, \overrightarrow{O B}=6 \vec{a}+5 \vec{b}$$ and $$\overrightarrow{O C}=3 \vec{b}$$, where $$O$$ is the origin. If the area of the parallelogram with adjacent sides $$\overrightarrow{O A}$$ and $$\overrightarrow{O C}$$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $$O A B C$$ is equal to:

Let $$\overrightarrow{\mathrm{a}}=4 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=11 \hat{i}-\hat{j}+\hat{k}$$ and $$\overrightarrow{\mathrm{c}}$$ be a vector such that $$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}})$$. If $$(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670$$, then $$|\vec{c}|^2$$ is equal to: