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

$|\vec{a}||\vec{b}||\vec{c}|=14$ <br/><br/> $$ \begin{aligned} & \vec{a} \wedge \vec{b}=\vec{b} \wedge \vec{c}=\vec{c} \wedge \vec{a}=\theta=\frac{2 \pi}{3} \\\\ & \vec{a} \cdot \vec{b}=-\frac{1}{2}|\vec{a}||\vec{b}| \\\\ & \vec{b} \cdot \vec{c}=-\frac{1}{2}|\vec{b}||\vec{c}| \\\\ & \vec{c} \cdot \vec{a}=-\frac{1}{2}|\vec{c}||\vec{a}| \end{aligned} $$ <br/><br/> Now, <br/><br/> $$ \begin{aligned} & \begin{aligned} &(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b}) \\\\ &=168 \quad\quad...(i)\\\\ &(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c})-(\vec{a} \cdot \vec{c})|\vec{b}|^{2} \\\\ &= \frac{1}{4}|\vec{b}|^{2}|\vec{a}||\vec{c}|+\frac{1}{2}|\vec{a}||\vec{b}|^{2}|\vec{c}| \\\\ &= \frac{3}{4}|\vec{a}||\vec{b}|^{2}|\vec{c}| \quad\quad...(ii) \end{aligned} \end{aligned} $$ <br/><br/> Similarly $(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})=\frac{3}{4}|\vec{a}||\vec{b}||\vec{c}|^{2} \quad\quad...(iii)$ <br/><br/> $(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=\frac{3}{4}|\vec{a}|^{2}|\vec{b}||\vec{c}| \quad\quad...(iv)$ <br/><br/> Substitute (ii), (iii), (iv) in (i) <br/><br/> $\frac{3}{4}|\vec{a}||\vec{b}||\vec{c}|[|\vec{a}|+|\vec{b}|+|\vec{c}|]=168$ <br/><br/> $\frac{3}{4} \times 14[|\vec{a}|+|\vec{b}|+|\vec{c}|]=168$ <br/><br/> $|\vec{a}|+|\vec{b}|+|\vec{c}|=16$