Let $\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}$ three coplanar concurrent vectors such that angles between any two ofthem is same. If the product of their magnitudes is 14 and $(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})+(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})\cdot (\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})+(\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})=168$ then $\left|\overrightarrow{\mathrm{a}}\right|+\left|\overrightarrow{\mathrm{b}}\right|+\left|\overrightarrow{\mathrm{c}}\right|$ is equal to :

$$\begin{gathered} \left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|\left|\overrightarrow{\mathrm{c}}\right|=14 \\ \overrightarrow{\mathrm{a}}\wedge \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}}\wedge \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}}\wedge \overrightarrow{\mathrm{a}}=\mathrm{\theta }=\frac{2\mathrm{\pi }}{3} \end{gathered}$$
So $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}=-\frac{1}{2}\mathrm{ab},\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}}=-\frac{1}{2}\mathrm{bc},\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{c}}=-\frac{1}{2}\mathrm{ac}$
(let)
$\begin{array}{l}(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})\cdot (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}})(\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}})-(\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{c}})(\overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{b}})\\ =\frac{1}{4}{\mathrm{ab}}^2\mathrm{c}+\frac{1}{2}{\mathrm{ab}}^2\mathrm{c}=\frac{3}{4}{\mathrm{ab}}^2\mathrm{c}\end{array}$
Similarly
$\begin{array}{l}(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})\cdot (\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})=\frac{3}{4}{\mathrm{abc}}^2\\ (\overrightarrow{\mathrm{c}}\times \overrightarrow{\mathrm{a}})\cdot (\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})=\frac{3}{4}{\mathrm{a}}^2\mathrm{bc}\\ 168=\frac{3}{4}\mathrm{abc}(\mathrm{a}+\mathrm{b}+\mathrm{c})\end{array}$
So, $(\mathrm{a}+\mathrm{b}+\mathrm{c})=16$