Let the vectors $\overrightarrow{AB} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{AC} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ be two sides of a triangle $ABC$. If $G$ is the centroid of $\triangle ABC$, then $\frac{27}{7}(\overrightarrow{AG})^2 + 5 =$

If the position vectors of two points $A$ and $B$ are $\vec{a} + 3\vec{b}$ and $\vec{a} - 2\vec{b}$ respectively,find the position vector of the point that divides $AB$ in the ratio $2:5$.

Classify the following measure as a scalar or a vector:
$10 \text{ kg}$

If $|\overline{a}|=2, |\overline{b}|=3, |\overline{c}|=5$ and each of the angles between the vectors $\overline{a}$ and $\overline{b}$,$\overline{b}$ and $\overline{c}$,and $\overline{c}$ and $\overline{a}$ is $60^{\circ}$,then the value of $|\overline{a}+\overline{b}+\overline{c}|$ is

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 $AG^2+BG^2+CG^2=$

Find the vector $\vec{c}$ which is in the direction of the internal angle bisector of the vectors $\vec{a} = 7\hat{i} - 4\hat{j} - 4\hat{k}$ and $\vec{b} = -2\hat{i} - \hat{j} + 2\hat{k}$ with $|\vec{c}| = 5\sqrt{6}$.