$\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are three vectors of equal magnitude. The angle between each pair of vectors is $\mathrm{\pi }/3$ such that $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|=\sqrt{6}$then $\left|\overrightarrow{\mathrm{a}}\right|$ is equal to

$\begin{array}{l}|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}{|}^2=\sqrt{6}\\ squaring onboth sides\\ \text{ ⇒}|\overrightarrow{\mathrm{a}}{|}^2+|\overrightarrow{\mathrm{b}}{|}^2+|\overrightarrow{\mathrm{c}}{|}^2+2(\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}.\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}.\overrightarrow{\mathrm{a}})=6\to (1)\\ Given that \left|\overrightarrow{\mathrm{a}}\right|=\left|\overrightarrow{\mathrm{b}}\right|=\left|\overrightarrow{\mathrm{c}}\right|\text{ and }\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|\cdot \cos \frac{\mathrm{\pi }}{3}\\ \text{ ⇒ }\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=\frac{1}{2}|\overrightarrow{\mathrm{a}}{|}^2\\ \therefore from \left(1\right) 3|\overrightarrow{\mathrm{a}}{|}^2+3|\overrightarrow{\mathrm{a}}{|}^2=6\\ \Rightarrow |\overrightarrow{\mathrm{a}}{|}^2=1\text{ ⇒ }|\overrightarrow{\mathrm{a}}|=1\end{array}$