Find the distance between the lines $l_{1}$ and $l_{2}$ given by $\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$ and $\vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+6 \hat{k})$.

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,then the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ is . . . . . . . (in $/2$)

$A$ hall has a square floor of dimension $10 \, m \times 10 \, m$ and vertical walls. If the angle $GPH$ between the diagonals $AG$ and $BH$ is $\cos^{-1} \frac{1}{5}$,then the height of the hall (in $meters$) is:

Let $\vec{a}, \vec{b}, \vec{c}$ be three coplanar concurrent vectors such that the angle between any two of them is the same. If the product of their magnitudes is $14$ and $(\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) + (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) + (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = 168$,then $|\vec{a}| + |\vec{b}| + |\vec{c}|$ is equal to:

Two adjacent sides of a triangle are represented by the vectors $\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{b} = 2\sqrt{3}\hat{i} - 2\sqrt{3}\hat{j} + \sqrt{3}\hat{k}$. Then the least angle of the triangle and the perimeter of the triangle are respectively:

Let $\overrightarrow{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\overrightarrow{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$. If $\overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{r}$,$\overrightarrow{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$ and $\overrightarrow{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$,where $\alpha \in R$,then the value of $\alpha + |\overrightarrow{r}|^{2}$ is equal to: