Find the area of a parallelogram whose adjacent sides are given by the vectors $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.

Let $\vec{a}=-\hat{i}+2\hat{j}+2\hat{k}$, $\vec{b}=8\hat{i}+7\hat{j}-3\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a}\times\vec{c}=\vec{b}$. If $\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k})=4$, then $|\vec{a}+\vec{c}|^{2}$ is equal to:

Let $\overrightarrow{a}=3 \hat{i}+2 \hat{j}+\hat{k}$,$\overrightarrow{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\overrightarrow{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$ and $(\overrightarrow{a}-\overrightarrow{b}+\hat{i}) \cdot \overrightarrow{c}=-3$. Then $|\overrightarrow{c}|^2$ is equal to . . . . . . .

If $a = i - 2j + 3k$ and $b = 3i + j + 2k,$ then the unit vector perpendicular to $a$ and $b$ is

Let $\vec{a}=\hat{i}+\alpha \hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}-\alpha \hat{j}+\hat{k}$. If the area of the parallelogram whose adjacent sides are represented by the vectors $\vec{a}$ and $\vec{b}$ is $8 \sqrt{3}$ square units,then $\vec{a} \cdot \vec{b}$ is equal to ....... .

Given $\bar{a} = 2\bar{i} + \bar{j} - 2\bar{k}$ and $\bar{b} = \bar{i} + \bar{j}$. If $\bar{c}$ is a vector such that $\bar{a} \cdot \bar{c} = |\bar{c}|$,$|\bar{c} - \bar{a}| = 2\sqrt{2}$,and the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$ is $30^{\circ}$,then the value of $|(\bar{a} \times \bar{b}) \times \bar{c}|^2$ is