MathematicsHard202×since 2002Q5830Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,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 (a⃗×b⃗)⋅(b⃗×c⃗)+(b⃗×c⃗)⋅(c⃗×a⃗)+(c⃗×a⃗)⋅(a⃗×b⃗)=168(\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(a×b)⋅(b×c)+(b×c)⋅(c×a)+(c×a)⋅(a×b)=168, then ∣a⃗∣+∣b⃗∣+∣c⃗∣|\vec{a}|+|\vec{b}|+|\vec{c}|∣a∣+∣b∣+∣c∣ is equal to :A10B14C16D18Check answerSkip