The area of the parallelogram,whose diagonals are $\vec{d}_1 = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{d}_2 = \hat{i} + 3 \hat{j} - \hat{k}$,is equal to

Let $\bar{a}$,$\bar{b}$,and $\bar{c}$ be unit vectors. Suppose that $\bar{a} \cdot \bar{b} = \bar{a} \cdot \bar{c} = 0$ and the angle between $\bar{b}$ and $\bar{c}$ is $\frac{\pi}{6}$. Then $\bar{a}$ is equal to:

The position vectors of the points $A, B, C$ are $\hat{i}+2\hat{j}-\hat{k}, \hat{i}+\hat{j}+\hat{k}$,and $2\hat{i}+3\hat{j}+2\hat{k}$ respectively. If $A$ is chosen as the origin,then the cross product of the position vectors of $B$ and $C$ is:

The unit vector perpendicular to the vector $\hat{i}-2 \hat{j}+3 \hat{k}$ and coplanar with the vectors $\hat{i}+\hat{j}+\hat{k}$ and $2 \hat{i}-\hat{j}-\hat{k}$ is

Let $\overrightarrow{c}$ be a vector perpendicular to the vectors $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k}.$ If $\overrightarrow{c}\cdot(\hat{i}+\hat{j}+3\hat{k})=8,$ then the value of $\overrightarrow{c}\cdot(\overrightarrow{a}\times\overrightarrow{b})$ is equal to ...... .

Let $\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}$,$\vec{b}=-\hat{i}+\hat{k}$,and $\vec{c}=\beta \hat{j}-\hat{k}$,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 with diagonals $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is $\frac{\sqrt{21}}{2}$,be $(\alpha_1, \beta_1)$ and $(\alpha_2, \beta_2)$. Then $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ is equal to