If $a, b, c$ are non-collinear vectors such that for some scalars $x, y, z,$ $xa + yb + zc = 0,$ then

If a point $C$ divides the line segment joining the points with the position vectors $2 \hat{i}-3 \hat{j}+2 \hat{k}$ and $3 \hat{i}-\hat{j}-2 \hat{k}$ in the ratio $2: 3$,then the distance of $C$ from the point with position vector $2 \hat{i}-\hat{j}+\hat{k}$ is

Let $A(3 \hat{i}+\hat{j}-\hat{k})$ and $B(13 \hat{i}-4 \hat{j}+9 \hat{k})$ be two points on a line $L$. $C$ and $D$ are points on $L$ on either side of $A$ at distances of $9$ and $6$ units respectively, and $C$ lies between $A$ and $B$. Then the position vectors of $C$ and $D$ are respectively:

If the position vectors of three points are $a$,$b$,and $(3a - 2b)$,then these points are .....

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 $\vec{OD} =$

If $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$,then $\vec{a} + \vec{b} = \dots$