Let $ABCDEF$ be a regular hexagon with the vertices $A, B, C, D, E$ and $F$ in counter-clockwise order. Then,the vector $\vec{AB} + \vec{BC}$ is parallel to:

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $|\vec{a}|=3, |\vec{b}|=5, |\vec{c}|=7$,then $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ does not exceed:

Let $\overline{OA}=\overline{a}, \overline{OB}=\overline{b}$. If the vector along the angle bisector of $\angle AOB$ is given by $x \frac{\overline{a}}{|\overline{a}|}+y \frac{\overline{b}}{|\overline{b}|}$,then which of the following is true?

Let $\overrightarrow{OA} = \hat{i} - 3\hat{j} + \hat{k}$, $\overrightarrow{OB} = \hat{i} + 3\hat{j} - 2\hat{k}$, and $\overrightarrow{OC} = 4\hat{i} + 3\hat{j} + 5\hat{k}$ be the position vectors of three points $A$, $B$, and $C$. Let $P$ be the point which divides $AB$ in the ratio $2:1$. If $l, m, n$ are the direction cosines of the vector $\overrightarrow{PC}$, then $l + 3m + 2n =$

If $\bar{a}, \bar{b}, \bar{c}$ are three unit vectors such that $|\bar{a}+\bar{b}|^2+|\bar{a}+\bar{c}|^2=8$,then $|\bar{a}+3\bar{b}|^2+|\bar{a}+3\bar{c}|^2=$

Consider the following statements :
Statement $(I)$ : If either $|\vec{a}|=0$ or $|\vec{b}|=0$,then $\vec{a} \cdot \vec{b}=0$.
Statement $(II)$ : If $\vec{a} \times \vec{b}=\vec{0}$,then $\vec{a}$ is perpendicular to $\vec{b}$.
Which of the following is correct?