Given $|\vec{a}|=\sqrt{3}$,$|\vec{b}|=5$,$\vec{b} \cdot \vec{c}=10$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$. If $\vec{a}$ is perpendicular to $\vec{b} \times \vec{c}$,then the value of $|\vec{a} \times(\vec{b} \times \vec{c})|$ is

$\vec{r}$ is a vector perpendicular to the plane determined by the vectors $2 \hat{i}-\hat{j}$ and $\hat{j}+2 \hat{k}$. If the magnitude of the projection of $\vec{r}$ on the vector $2 \hat{i}+\hat{j}+2 \hat{k}$ is $1$, then $|\vec{r}|=$

If $(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + \lambda \hat{j} + \mu \hat{k}) = \vec{0}$,then $\lambda$ and $\mu$ are respectively:

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

Let $L_1: \frac{x+2}{5}=\frac{y-3}{2}=\frac{z-6}{1}$ and $L_2: \frac{x-3}{4}=\frac{y+2}{3}=\frac{z-3}{5}$ be the given lines. Then the unit vector perpendicular to both $L_1$ and $L_2$ is

$A$ unit vector perpendicular to the plane containing the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$ is