Let vec{a}, vec{b}, vec{c} be three coplanar | vedclass

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(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are coplanar and the angle between any two is the same,the angle $	heta = \frac{2\pi}{3} = 120^\circ$.Let

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$16$

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(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are coplanar and the angle between any two is the same,the angle $	heta = \frac{2\pi}{3} = 120^\circ$.Let $a = |\vec{a}|, b = |\vec{b}|, c = |\vec{c}|$. We are given $abc = 14$.Using the vector identity $(\vec{u} 	imes \vec{v}) \cdot (\vec{w} 	imes \vec{z}) = (\vec{u} \cdot \vec{w})(\vec{v} \cdot \vec{z}) - (\vec{u} \cdot \vec{z})(\vec{v} \cdot \vec{w})$:$(\vec{a} 	imes \vec{b}) \cdot (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c}) - (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{b}) = (a b \cos 120^\circ)(b c \cos 120^\circ) - (a c \cos 120^\circ)(b^2) = (a b \cdot -\frac{1}{2})(b c \cdot -\frac{1}{2}) - (a c \cdot -\frac{1}{2})(b^2) = \frac{1}{4} a b^2 c + \frac{1}{2} a b^2 c = \frac{3}{4} a b^2 c$.Similarly,$(\vec{b} 	imes \vec{c}) \cdot (\vec{c} 	imes \vec{a}) = \frac{3}{4} a b c^2$ and $(\vec{c} 	imes \vec{a}) \cdot (\vec{a} 	imes \vec{b}) = \frac{3}{4} a^2 b c$.Summing these,we get $\frac{3}{4} a b c (a + b + c) = 168$.Substituting $abc = 14$: $\frac{3}{4} \cdot 14 \cdot (a + b + c) = 168$.$\frac{21}{2} (a + b + c) = 168 \implies a + b + c = 168 \cdot \frac{2}{21} = 8 \cdot 2 = 16$.

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