Find the sum of the vectors $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$,$\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}$,and $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$.

$P$ and $Q$ are the points of trisection of the line segment $AB$. If $2 \hat{i}-5 \hat{j}+3 \hat{k}$ and $4 \hat{i}+\hat{j}-6 \hat{k}$ are the position vectors of $A$ and $B$ respectively, then the position vector of the point which divides $PQ$ in the ratio $2:3$ is

$ABCD$ is a parallelogram and $P$ is the mid-point of the side $AD$. The line $BP$ meets the diagonal $AC$ in $Q$. Then, the ratio of $AQ:QC$ is equal to

The vectors $a = 2\hat{i} + 3\hat{j} + 6\hat{k}$ and $b$ are collinear and $|b| = 21$,then $b =$

If $a$ and $b$ represent two non-collinear vectors, the equation $r = ta + (1-t)b$ represents

Let $\vec{OA} = -4\hat{i} + 3\hat{k}$ and $\vec{OB} = 14\hat{i} + 2\hat{j} - 5\hat{k}$. If $\vec{OD}$ bisects $\angle AOB$ and $|\vec{OD}| = \sqrt{6}$,then $\vec{OD} =$