Let the position vectors of two points $A$ and $B$ be $\vec{a}+\vec{b}+\vec{c}$ and $\vec{a}-2\vec{b}+3\vec{c}$,respectively. If the points $P$ and $Q$ divide $AB$ in the ratio $1:3$ internally and externally respectively,then $3|AB|=$

$ABCD$ is a tetrahedron. $\bar{i}-2\bar{j}+3\bar{k}$, $-2\bar{i}+\bar{j}+3\bar{k}$, and $3\bar{i}+2\bar{j}-\bar{k}$ are the position vectors of the points $A, B, C$ respectively. $-\bar{i}+2\bar{j}-3\bar{k}$ is the position vector of the centroid of the triangular face $BCD$. If $G$ is the centroid of the tetrahedron, then $GD=$

If $\vec{a}=-2 \hat{i}+9 \hat{j}-6 \hat{k}$ and $\vec{b}=t \hat{i}-2 \hat{j}+6 \hat{k}$ are vectors such that $|\vec{a}+\vec{b}|=25$,then the sum of the values of $t$ is

If $D, E, F$ are the midpoints of the sides $BC, CA, AB$ of a triangle $ABC$ respectively,find the position vector of the centroid of $\Delta DEF$,given that the position vectors of $A, B, C$ are $i + j, j + k, k + i$ respectively.

The points with position vectors $\bar{a}+\bar{b}$,$\bar{a}-\bar{b}$,and $\bar{a}+k\bar{b}$ are collinear:

Let a vector $\alpha \hat{i}+\beta \hat{j}$ be obtained by rotating the vector $\sqrt{3} \hat{i}+\hat{j}$ by an angle $45^{\circ}$ about the origin in the counterclockwise direction in the first quadrant. Then the area of the triangle having vertices $(\alpha, \beta), (0, \beta)$ and $(0,0)$ is equal to