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 the components of $\vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ respectively,are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})$,then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Let $\overline{A}, \overline{B}, \overline{C}$ be vectors of lengths $3$ units,$4$ units,and $5$ units respectively. If $\overline{A}$ is perpendicular to $\overline{B}+\overline{C}$,$\overline{B}$ is perpendicular to $\overline{C}+\overline{A}$,and $\overline{C}$ is perpendicular to $\overline{A}+\overline{B}$,then the length of vector $\overline{A}+\overline{B}+\overline{C}$ is

In a $\triangle ABC$,$|CB|=a$,$|CA|=b$,$|AB|=c$ and $CD$ is the median through the vertex $C$. Then,$CA \cdot CD=$

The figure formed by the four points $i + j - k$,$2i + 3j$,$3i + 5j - 2k$,and $k - j$ is:

If vectors $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$,and $\bar{c}=-3 \hat{i}+\hat{j}+2 \hat{k}$ are such that $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then $\lambda=$