Given three vectors $\bar{a}, \bar{b}, \bar{c}$,two of which are collinear. If $\bar{a}+\bar{b}$ is collinear with $\bar{c}$ and $\bar{b}+\bar{c}$ is collinear with $\bar{a}$,and $|\bar{a}|=|\bar{b}|=|\bar{c}|=\sqrt{2}$,then $\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}=$

If $\hat{a}, \hat{b},$ and $\hat{c}$ are unit vectors satisfying $\hat{a} - \sqrt{3}\hat{b} + \hat{c} = \vec{0},$ then the angle between the vectors $\hat{a}$ and $\hat{c}$ is

If $S$ is the circumcentre, $O$ is the orthocentre and $G$ is the centroid of a triangle $ABC$, then match the items of the List-$I$ with those of the items of List-$II$ given below.
| List-$I$ | List-$II$ |
| :--- | :--- |
| $(i)$ $\vec{SA} + \vec{SB} + \vec{SC}$ | $(A)$ $2\vec{OS}$ |
| (ii) $\vec{GA} + \vec{GB} + \vec{GC}$ | $(B)$ $\frac{2}{3}\vec{OS}$ |
| (iii) $\vec{OA} + \vec{OB} + \vec{OC}$ | $(C)$ $\vec{0}$ |
| (iv) $\vec{OG}$ | $(D)$ $\vec{SO}$ |
| | $(E)$ $\vec{OS}$ |

The position vectors of the vertices $A, B$ and $C$ of a triangle are $2 \hat{i}-3 \hat{j}+3 \hat{k}$,$2 \hat{i}+2 \hat{j}+3 \hat{k}$ and $-\hat{i}+\hat{j}+3 \hat{k}$ respectively. Let $l$ denote the length of the angle bisector $AD$ of $\angle BAC$,where $D$ is on the line segment $BC$. Then $2 l^2$ equals:

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors,$|\overline{a}|=2, |\overline{b}|=4, |\overline{c}|=1$,$|\overline{b} \times \overline{c}|=\sqrt{15}$ and $\overline{b}=2 \overline{c}+\lambda \overline{a}$,then the value of $\lambda$ is

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $\overline{a} \cdot(\overline{b}+\overline{c})+\overline{b} \cdot(\overline{c}+\overline{a})+\overline{c} \cdot(\overline{a}+\overline{b})=0$ and $|\overline{a}|=1$,$|\overline{b}|=8$ and $|\overline{c}|=4$,then $|\overline{a}+\overline{b}+\overline{c}|$ has the value