Let $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}, \vec{c}=\hat{i}-\hat{j}$ and $\vec{d}=\hat{i}+\hat{j}+x \hat{k}$. If $(\vec{a} \times \vec{b}) \times \vec{c}$ is perpendicular to $\vec{d}$, then $x=$

Let $\bar{a}=\bar{i}-\bar{j}+\bar{k}, \bar{b}=\bar{i}-2\bar{j}-2\bar{k}, \bar{c}=6\bar{i}+3\bar{j}-2\bar{k}$ be three vectors. If $\bar{d}$ is a vector perpendicular to both $\bar{a}$ and $\bar{b}$, and $|\bar{d} \times \bar{c}|=14$, then $|\bar{d} \cdot \bar{c}|=$

Let $\overline{a}, \overline{b}, \overline{c}$ be three vectors such that $|\overline{a}|=\sqrt{3}$,$|\overline{b}|=5$,$\overline{b} \cdot \overline{c}=10$ and the angle between $\overline{b}$ and $\overline{c}$ is $\frac{\pi}{3}$. If $\overline{a}$ is perpendicular to the vector $\overline{b} \times \overline{c}$,then $|\overline{a} \times(\overline{b} \times \overline{c})|$ is equal to

Consider the lines $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$. Then the unit vector perpendicular to both $L_1$ and $L_2$ is:

If $\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{b} = \hat{i} + 3\hat{j} + 5\hat{k}$ and $\overrightarrow{c} = 7\hat{i} + 9\hat{j} + 11\hat{k}$,then the area of the parallelogram having diagonals $\overrightarrow{a} + \overrightarrow{b}$ and $\overrightarrow{b} + \overrightarrow{c}$ is:

Find a unit vector coplanar with $i + 2j + k$ and $i + j + 2k$ and perpendicular to $i + j + k$.