If $ABCDEF$ is a regular hexagon with $\vec{AB} = \vec{a}$ and $\vec{BC} = \vec{b}$,then $\vec{CE}$ equals to

Using vectors,find the value of $k$ such that the points $(k,-10,3), (1,-1,3)$ and $(3,5,3)$ are collinear.

If $\hat{i}$ is the position vector of the centroid $G$ of triangle $ABC$ and $2\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}-4\hat{k}$ are respectively the position vectors of its vertices $A$ and $B$,then $AG^2+BG^2+CG^2=$

If $\overline{a} = \bar{i} - 2\bar{j} + 2\bar{k}$ and $\overline{b} = 9\bar{i} + 6\bar{j} - 18\bar{k}$ are two vectors,then $\frac{\text{Projection of } \overline{b} \text{ on } \overline{a}}{\text{Projection of } \overline{a} \text{ on } \overline{b}} = $

Show that the points $A (-2 \hat{i}+3 \hat{j}+5 \hat{k})$,$B (\hat{i}+2 \hat{j}+3 \hat{k})$ and $C (7 \hat{i}-\hat{k})$ are collinear.

If $\alpha, \beta, \gamma$ are distinct real numbers and $\alpha+\beta+\gamma \neq 0$,then the points with position vectors $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}, \beta \hat{i}+\gamma \hat{j}+\alpha \hat{k}$ and $\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k}$ are