The shortest distance between the lines $r = 3i + 5j + 7k + \lambda(i + 2j + k)$ and $r = -i - j - k + \mu(7i - 6j + k)$ is

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

If $\vec{a} = 2\hat{i} - \hat{j} - 2\hat{k}$ and $\vec{b} = 6\hat{i} + 2\hat{j} - 3\hat{k}$ are two vectors,and we consider a vector $\vec{c} = \vec{a} + t\vec{b}$,find the value of $t$ such that the magnitude $|\vec{c}|$ is minimum.

The projection of the vector $2i + j - 3k$ on the vector $i - 2j + k$ is:

The angle between the pair of lines with direction ratios $(1, 1, 2)$ and $(\sqrt{3} - 1, -\sqrt{3} - 1, 4)$ is ......... $^o$.

Three vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy the condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If $|\vec{a}|=1, |\vec{b}|=3, |\vec{c}|=4$,then find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$.