If the position vectors of the vertices of a triangle are $a, b, c$,then find the sum of the vectors from the vertices to the centroid.

If the vectors $x \hat{i}-3 \hat{j}+7 \hat{k}$ and $\hat{i}+y \hat{j}-z \hat{k}$ are collinear,then the value of $\frac{x y^2}{z}$ is equal to:

If the position vectors of $A$ and $B$ are respectively $(1, 1, 0)$ and $(0, 1, 1)$,then $\overrightarrow{AB} =$

If $a, b, c$ are the position vectors of three collinear points,then the existence of scalars $x, y, z$ (not all zero) is such that:

Let $\overrightarrow{OA}=\hat{i}+2 \hat{j}-2 \hat{k}$ and $\overrightarrow{OB}=-2 \hat{i}-3 \hat{j}+6 \hat{k}$ be the position vectors of two points $A$ and $B$. If $C$ is a point on the bisector of $\angle AOB$ and $OC=\sqrt{42}$,then $\overrightarrow{OC}=$

Find a vector in the direction of vector $\vec{a} = \hat{i} - 2\hat{j}$ that has magnitude $7$ units.