JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Vector Algebra

Since the angles between any two vectors are equal and they are coplanar, each angle is $\theta =\frac{2\pi }{3}=120^{\circ }$. Thus $\vec{a}\cdot \vec{b}=-\frac{1}{2}ab$, $\vec{b}\cdot \vec{c}=-\frac{1}{2}bc$, $\vec{a}\cdot \vec{c}=-\frac{1}{2}ac$. Using the vector identity $(\vec{a}\times \vec{b})\cdot (\vec{b}\times \vec{c})=(\vec{a}\cdot \vec{b})(\vec{b}\cdot \vec{c})-(\vec{a}\cdot \vec{c})(\vec{b}\cdot \vec{b})=\frac{1}{4}a{b}^{2}c+\frac{1}{2}a{b}^{2}c=\frac{3}{4}a{b}^{2}c$. Similarly, the other two terms give $\frac{3}{4}ab{c}^2$ and $\frac{3}{4}{a}^{2}bc$. Summing: $168=\frac{3}{4}abc(a+b+c)$. Since $abc=14$, we have $168=\frac{3}{4}\cdot 14(a+b+c)$, so $a+b+c=16$.