Let $\vec{a}, \vec{b}, \vec{c}$ three coplanar concurrent vectors such that angles between any two ofthem 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 :

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Detailed Solution

$| \vec{a} | | \vec{b} | | \vec{c} | = 14 \vec{a} \land \vec{b} = \vec{b} \land \vec{c} = \vec{c} \land \vec{a} = θ = \frac{2 π}{3}$
So $\vec{a} \cdot \vec{b} = - \frac{1}{2} ab, \vec{b} \cdot \vec{c} = - \frac{1}{2} bc, \vec{a} \cdot \vec{c} = - \frac{1}{2} ac$
(let)
$\begin{array}{l} (\vec{a} \times \vec{b}) \cdot (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{b}) (\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c}) (\vec{b} \cdot \vec{b}) \\ = \frac{1}{4} {ab}^{2} c + \frac{1}{2} {ab}^{2} c = \frac{3}{4} {ab}^{2} c \end{array}$
Similarly
$\begin{array}{l} (\vec{b} \times \vec{c}) \cdot (\vec{c} \times \vec{a}) = \frac{3}{4} {abc}^{2} \\ (\vec{c} \times \vec{a}) \cdot (\vec{a} \times \vec{b}) = \frac{3}{4} a^{2} bc \\ 168 = \frac{3}{4} abc (a + b + c) \end{array}$
So, $(a + b + c) = 16$