Let → a , → b , → c be three coplanar concurrent vectors such that...

Given, product of magnitudes is $14$

So, \left|\overset{\to}{a}\right|\left|\overset{\to}{b}\right|\left|\overset{\to}{c}\right| = 14$</p><p>Also given angles between any two of them is same</p><p>So, angle between$\overset{\to}{a} \text{ and } \overset{\to}{b} = \overset{\to}{b} \text{ and } \overset{\to}{c} = \overset{\to}{c} \text{ and } \overset{\to}{a} = \theta = \dfrac{2\pi}{3}$</p><p>So,$\overset{\to}{a} . \overset{\to}{b} = - \dfrac{1}{2}ab, \overset{\to}{b} . \overset{\to}{c} = - \dfrac{1}{2}bc \text{ and } \overset{\to}{a} . \overset{\to}{c} = - \dfrac{1}{2}ac$</p><p>Now solving \left(\overset{\to}{a} \times \overset{\to}{b}\right) \cdot \left(\overset{\to}{b} \times \overset{\to}{c}\right) = \left(\overset{\to}{a} \cdot \overset{\to}{b}\right)\left(\overset{\to}{b} \cdot \overset{\to}{c}\right) - \left(\overset{\to}{a} \cdot \overset{\to}{c}\right)\left(\overset{\to}{b} \cdot \overset{\to}{b}\right)$

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

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

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

Similarly for

\left(\overset{\to}{b} \times \overset{\to}{c}\right) \cdot \left(\overset{\to}{c} \times \overset{\to}{a}\right) = \dfrac{3}{4}ab{c}^{2} . . . . . . . . . \left(2\right)$</p><p>And \left(\overset{\to}{c} \times \overset{\to}{a}\right) \cdot \left(\overset{\to}{a} \times \overset{\to}{b}\right) = \dfrac{3}{4}{a}^{2}bc . . . . . . \left(3\right)$

Now adding equation \left(1\right), \left(2\right) \text{ and } \left(3\right)$we get,</p><p>\left(\overset{\to}{a} \times \overset{\to}{b}\right) \cdot \left(\overset{\to}{b} \times \overset{\to}{c}\right) + \left(\overset{\to}{b} \times \overset{\to}{c}\right) \cdot \left(\overset{\to}{c} \times \overset{\to}{a}\right) + \left(\overset{\to}{c} \times \overset{\to}{a}\right) \cdot \left(\overset{\to}{a} \times \overset{\to}{b}\right) = 168$

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

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

So, \left(a + b + c\right) = 16