If $a \perp b$ and $(a+b) \perp (a+mb)$,then $m$ is equal to

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) $\vec{OG}$ | $(D)$ $\vec{SO}$ |
| | $(E)$ $\vec{OS}$ |

If $12 \hat{i}-12 \hat{j}-18 \hat{k}$,$-3 \hat{i}-6 \hat{j}-9 \hat{k}$ and $3 \hat{i}+3 \hat{j}-24 \hat{k}$ are the position vectors of the vertices $A, B$ and $C$ respectively of $\triangle ABC$,then the position vector of the incentre of $\triangle ABC$ is

$A, B, C, D$ 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 $\triangle ABC$.

$ABCD$ is a parallelogram and $P$ is a point on the segment $AD$ dividing it internally in the ratio $3:1$. If the line $BP$ meets the diagonal $AC$ in $Q$,then $AQ:QC$ equals

If $|\bar{a}|=\sqrt{26}$,$|\bar{b}|=7$,and $|\bar{a} \times \bar{b}|=35$,then $\bar{a} \cdot \bar{b}=$