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

Let $L_1: \overrightarrow{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in R$,$L_2: \overrightarrow{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in R$,and $L_3: \overrightarrow{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in R$ be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$. Then the point which lies on $L_3$ is

$x, y, z$ are three vectors each of magnitude $\sqrt{2}$ and each making an angle $60^{\circ}$ with one another. If $a=x \times(y \times z), b=y \times(z \times x)$, $c=x \times y$, then $x=$

$A$ unit vector perpendicular to both $i+j+k$ and $2i+j+3k$ is

The area of the parallelogram,whose diagonals are $\vec{d}_1 = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{d}_2 = \hat{i} + 3 \hat{j} - \hat{k}$,is equal to

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: