$ABC$ is an equilateral triangle. Length of each side is $a$ and centroid is point $O$. Find $\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=.......$

Statement $I:$ If three forces $\vec{F}_{1}, \vec{F}_{2}$ and $\vec{F}_{3}$ are represented by three sides of a triangle and $\overrightarrow{{F}}_{1}+\overrightarrow{{F}}_{2}=-\overrightarrow{{F}}_{3}$, then these three forces are concurrent forces and satisfy the condition for equilibrium.

Statement $II:$ A triangle made up of three forces $\overrightarrow{{F}}_{1}, \overrightarrow{{F}}_{2}$ and $\overrightarrow{{F}}_{3}$ as its sides taken in the same order, satisfy the condition for translatory equilibrium.

In the light of the above statements, choose the most appropriate answer from the options given below:

Given that $\overrightarrow A + \overrightarrow B + \overrightarrow C= 0$ out of three vectors two are equal in magnitude and the magnitude of third vector is $\sqrt 2 $ times that of either of the two having equal magnitude. Then the angles between vectors are given by

If $\vec A$ and $\vec B$ are two non-zero vectors such that $\left| {\vec A + \vec B} \right| = \frac{{\left| {\vec A - \vec B} \right|}}{2}$ and $\left| {\vec A} \right| = 2\left| {\vec B} \right|$ then the angle between $\vec A$ and $\vec B$ is

The ratio of maximum and minimum magnitudes of the resultant of two vector $\vec a$ and $\vec b$ is $3 : 1$. Now $| \vec a |$  is equal to

A particle is simultaneously acted by two forces equal to $4\, N$ and $3 \,N$. The net force on the particle is