If the position vector of one end of a line segment $AB$ is $2i + 3j - k$ and the position vector of its midpoint is $3(i + j + k)$,then what is the position vector of the other end?

Let $3 \hat{i}+\hat{j}-\hat{k}$ be the position vector of a point $B$. Let $A$ be a point on the line which is passing through $B$ and parallel to the vector $2 \hat{i}-\hat{j}+2 \hat{k}$. If $|\overrightarrow{B A}|=18$,then the position vector of $A$ is

If the vectors $\hat{i}+2 \hat{j}+x \hat{k}$ and $y \hat{i}+6 \hat{j}+4 \hat{k}$ are collinear,then the values of $x$ and $y$ are respectively,

If $\vec{a}$ and $\vec{b}$ are the position vectors $(P.V.)$ of two points $A$ and $B$ respectively,and $C$ divides the line segment $AB$ in the ratio $2 : 1$,then the position vector of $C$ is:

If $\alpha, \beta, \gamma$ are distinct real numbers and $\alpha+\beta+\gamma \neq 0$,then the points with position vectors $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}, \beta \hat{i}+\gamma \hat{j}+\alpha \hat{k}$ and $\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k}$ are

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: