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

Given, product of magnitudes is $14$

So, $\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|=14$

Also given angles between any two of them is same

So, angle between $\overrightarrow{a} \& \overrightarrow{b}=\overrightarrow{b} \& \overrightarrow{c}=\overrightarrow{c} \& \overrightarrow{a}=\theta =\frac{2\pi }{3}$

So, $\overrightarrow{a}.\overrightarrow{b}=-\frac{1}{2}ab, \overrightarrow{b}.\overrightarrow{c}=-\frac{1}{2}bc \& \overrightarrow{a}.\overrightarrow{c}=-\frac{1}{2}ac$

Now solving $\left(\overrightarrow{a}\times \overrightarrow{b}\right)·\left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left(\overrightarrow{a}·\overrightarrow{b}\right)\left(\overrightarrow{b}·\overrightarrow{c}\right)-\left(\overrightarrow{a}·\overrightarrow{c}\right)\left(\overrightarrow{b}·\overrightarrow{b}\right)$

$=\left(\frac{-1}{2}ab\right)\left(\frac{-1}{2}bc\right)-\left(\frac{-1}{2}ac\right)\left(b^2\right)$

$=\frac{1}{4}a{b}^{2}c+\frac{1}{2}a{b}^{2}c$

$=\frac{3}{4}a{b}^{2}c .......\left(1\right)$

Similarly for 

$\left(\overrightarrow{b}\times \overrightarrow{c}\right)·\left(\overrightarrow{c}\times \overrightarrow{a}\right)=\frac{3}{4}ab{c}^2 .........\left(2\right)$

And $\left(\overrightarrow{c}\times \overrightarrow{a}\right)·\left(\overrightarrow{a}\times \overrightarrow{b}\right)=\frac{3}{4}{a}^{2}bc ......\left(3\right)$

Now adding equation $\left(1\right), \left(2\right) \& \left(3\right)$ we get,

$\left(\overrightarrow{a}\times \overrightarrow{b}\right)·\left(\overrightarrow{b}\times \overrightarrow{c}\right)+\left(\overrightarrow{b}\times \overrightarrow{c}\right)·\left(\overrightarrow{c}\times \overrightarrow{a}\right)+\left(\overrightarrow{c}\times \overrightarrow{a}\right)·\left(\overrightarrow{a}\times \overrightarrow{b}\right)=168$

$\Rightarrow \frac{3}{4}abc\left(a+b+c\right)=168$

$\Rightarrow \left(a+b+c\right)=\frac{168\times 4}{14\times 3}$

So, $\left(a+b+c\right)=16$