The area of the parallelogram determined by vectors $A = 2\hat{i} + \hat{j} - 3\hat{k}$ and $B = 12\hat{j} - 2\hat{k}$ is approximately:

If $\overrightarrow{A} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\overrightarrow{B} = -\hat{i} + 3\hat{j} + 4\hat{k}$,then the projection of $\overrightarrow{A}$ on $\overrightarrow{B}$ will be:

If $\vec{A} = 4\hat{i} + n\hat{j} - 2\hat{k}$ and $\vec{B} = 2\hat{i} + 3\hat{j} + \hat{k}$,find the value of $n$ such that $\vec{A} \perp \vec{B}$.

The diagonals of a parallelogram are $2\,\hat{i}$ and $2\,\hat{j}$. What is the area of the parallelogram in square units?

Consider three vectors $A = \hat{i} + \hat{j} - 2\hat{k}$,$B = \hat{i} - \hat{j} + \hat{k}$,and $C = 2\hat{i} - 3\hat{j} + 4\hat{k}$. $A$ vector $X$ of the form $X = \alpha A + \beta B$ (where $\alpha$ and $\beta$ are scalars) is perpendicular to $C$. The ratio of $\alpha$ to $\beta$ is: (in $: 1$)

Given two vectors $\vec{A} = 3\hat{i} + \hat{j}$ and $\vec{B} = \hat{j} + 2\hat{k}$. If these two vectors represent the adjacent sides of a parallelogram,find the area of the parallelogram.