Three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy the condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}.$ Evaluate the quantity $\mu=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a},$ if $|\vec{a}|=1, |\vec{b}|=4$ and $|\vec{c}|=2.$

$A, B, C, D$ are four points in a plane with position vectors $\overline{a}, \overline{b}, \overline{c}, \overline{d}$ respectively such that $(\overline{a}-\overline{d}) \cdot(\overline{b}-\overline{c})=(\overline{b}-\overline{d}) \cdot(\overline{c}-\overline{a})=0$. Then the point $D$ is the $\dots$ of $\triangle ABC$.

If $a \cdot i = a \cdot (i + j) = a \cdot (i + j + k)$, then $a = $

Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be three non-zero vectors which are pairwise non-collinear. If $\vec{\alpha}+3 \vec{\beta}$ is collinear with $\vec{\gamma}$ and $\vec{\beta}+2 \vec{\gamma}$ is collinear with $\vec{\alpha}$, then $\vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}$ is

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{c} = \hat{j} - \hat{k}$ and a vector $\vec{b}$ be such that $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 3$. Then $|\vec{b}|$ equals?

If $a$ makes an acute angle with $b$,$r \cdot a = 0$ and $r \times b = c \times b$,then $r=$