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

If $a \times (b \times c) = 0,$ then

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

$a = 3i - 5j$ and $b = 6i + 3j$ are two vectors and $c$ is a vector such that $c = a \times b$,then $|a|:|b|:|c|$ is

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

If $\overrightarrow{u}=\overrightarrow{a}-\overrightarrow{b}$,$\overrightarrow{v}=\overrightarrow{a}+\overrightarrow{b}$,$|\overrightarrow{a}|=|\overrightarrow{b}|=2$,then $|\overrightarrow{u} \times \overrightarrow{v}|$ is equal to