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.

Let $\overline{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\overline{b}=\hat{i}+\hat{j}$. If $\overline{c}$ is a vector such that $\overline{a} \cdot \overline{c}=|\overline{c}|$,$|\overline{c}-\overline{a}|=2 \sqrt{2}$ and the angle between $(\overline{a} \times \overline{b})$ and $\overline{c}$ is $\frac{\pi}{6}$,then $|(\overline{a} \times \overline{b}) \times \overline{c}|$ is

If $2 \hat{i}+3 \hat{j}-4 \hat{k}$ and $-\hat{i}+2 \hat{j}+\hat{k}$ are the two diagonals of a parallelogram, then the area of the parallelogram in square units is

Let $\overrightarrow{a}=3\hat{i}-\hat{j}+2\hat{k}$,$\overrightarrow{b}=\overrightarrow{a}\times(\hat{i}-2\hat{k})$ and $\overrightarrow{c}=\overrightarrow{b}\times\hat{k}$. Then the projection of $\overrightarrow{c}-2\hat{j}$ on $\overrightarrow{a}$ is:

If $C$ is a given non-zero scalar and $\overline{A}$ and $\overline{B}$ are given non-zero vectors such that $\overline{A}$ is perpendicular to $\overline{B}$. If vector $\overline{X}$ is such that $\overline{A} \cdot \overline{X} = C$ and $\overline{A} \times \overline{X} = \overline{B}$,then $\overline{X}$ is given by:

If $\overline{a}=\hat{i}+4 \hat{j}+2 \hat{k}$,$\overline{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$,and $\overline{c}=2 \hat{i}-\hat{j}+4 \hat{k}$,then a vector $\bar{d}$ which is parallel to vector $\overline{a} \times \overline{b}$ and satisfies $\overline{c} \cdot \overline{d}=15$,is