Two adjacent sides of a parallelogram are given by vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$. Find the area of the parallelogram in square units.

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:

If $a=2\hat{i}+\hat{j}-3\hat{k}$,$b=\hat{i}-2\hat{j}+\hat{k}$,$c=-\hat{i}+\hat{j}-4\hat{k}$ and $d=\hat{i}+\hat{j}+\hat{k}$,then $|(a \times b) \times(c \times d)|=$

The area of the triangle,whose vertices are $A \equiv(1,-1,2)$,$B \equiv(2,1,-1)$ and $C \equiv(3,-1,2)$,is

Let $m$ be a vector of magnitude $\sqrt{3}$ and perpendicular to the vectors $\hat{i}+\hat{j}$ and $\hat{j}-\hat{k}$. Let $n$ be another vector of magnitude $2\sqrt{6}$ and perpendicular to the vectors $2\hat{i}-\hat{j}$ and $\hat{j}+2\hat{k}$. The area (in sq. units) of the triangle formed with $m$ and $n$ as sides is

Let $\overrightarrow{a} = \alpha \hat{i} + 3 \hat{j} - \hat{k}$,$\overrightarrow{b} = 3 \hat{i} - \beta \hat{j} + 4 \hat{k}$ and $\overrightarrow{c} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ where $\alpha, \beta \in \mathbb{R}$,be three vectors. If the projection of $\overrightarrow{a}$ on $\overrightarrow{c}$ is $\frac{10}{3}$ and $\overrightarrow{b} \times \overrightarrow{c} = -6 \hat{i} + 10 \hat{j} + 7 \hat{k}$,then the value of $\alpha + \beta$ is equal to: