The zero vector $(0,0,0)$ $........$

If $A(a)$, $B(b)$, and $C(c)$ are vertices of $\Delta ABC$. Point $D$ divides segment $BC$ internally in the ratio $2 : 1$. Point $E$ divides segment $AD$ internally in the ratio $1 : 2$, then the position vector of $E$ is

If a parallelogram is constructed on the vectors $\vec{a} = 3\vec{p} - \vec{q}$ and $\vec{b} = \vec{p} + 3\vec{q}$, where $|\vec{p}| = 3$, $|\vec{q}| = 2$ and the angle between $\vec{p}$ and $\vec{q}$ is $\pi/3$, then the ratio of the lengths of adjacent sides $|\vec{a}|$ and $|\vec{b}|$ of the parallelogram is:

The position vectors of the points $A$ and $B$ are $\vec{a}$ and $\vec{b}$ respectively. If the position vector of the point $C$ is $\frac{\vec{a}}{2} + \frac{\vec{b}}{3}$, then:

The direction cosines of the vector $\hat{i}-2\hat{j}+3\hat{k}$ are . . . . . . .

Let $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=-5 \hat{i}+7 \hat{j}$,and $\vec{c}=3 \hat{i}+y \hat{j}$ be three vectors such that $|\vec{a}-\vec{b}+\vec{c}|=\sqrt{141}$. If $y_1$ and $y_2$ are the values of $y$ satisfying the given condition,then $|y_1-y_2|=$