Three concurrent forces of the same magnitude are in equilibrium. What is the angle between the forces  Also name the triangle formed by the forces as sides

The value of the sum of two vectors $\overrightarrow A $ and $\overrightarrow B $ with $\theta $ as the angle between them is

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:

If a particle moves from point $P (2,3,5)$ to point $Q (3,4,5)$. Its displacement vector be

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

Mark the correct statement :-