$ABC$ is an equilateral triangle of side $‘ a ’ .$ Then $AB . BC + BC . CA + CA . AB =$

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$- \frac{3 a^{2}}{2}$

$- 2 a^{2}$

$- \frac{a^{2}}{2}$

$- a^{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} \vec{A B} + \vec{B C} + \vec{C A} = 0 \\ | \vec{A B} + \vec{B C} + \vec{C A} |^{2} = 0 \\ | \vec{A B} |^{2} + | \vec{B C} |^{2} + | \vec{C A} |^{2} + 2 (\vec{A B} \cdot \vec{B C} + \vec{B C} \cdot \vec{C A} + \vec{C A} \cdot \vec{A B}) = 0 \\ 2 (\vec{A B} \cdot \vec{B C} + \vec{B C} \cdot \vec{C A} + \vec{C A} \cdot \vec{A B}) = - (| \vec{A B} |^{2} + | \vec{B C} |^{2} + | \vec{C A} |^{2}) \\ = - (a^{2} + a^{2} + a^{2}) since, | \vec{A B} | = | \vec{B C} | = | \vec{C A} | = a \\ = - 3 a^{2} \\ \vec{A B} \cdot \vec{B C} + \vec{B C} \cdot \vec{C A} + \vec{C A} \cdot \vec{A B} = - 3 / 2 a^{2} \end{array}$