In $△ A B C, a sin (B - C) + b sin (C - A) + c sin (A - B)$ is equal to

0

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a + b + c

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a^{2} + b^{2} + c^{2}

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2 (a^{2} + b^{2} + c^{2})

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Solution

The correct option is A $0$
Use sine formula
$\frac{s i n A}{a} = \frac{s i n B}{b} = \frac{s i n C}{c} = k$
$∴ s i n A = a k$
$s i n B = b k$
$s i n C = c k$
Also
$s i n (A - B) = s i n A c o s B - c o s A s i n B$
$\Rightarrow s i n (A - B) = a k cos B - c o s A b k$
$= k (a c o s B - b cos A)$
Similarly
$s i n (B - C) = k (b c o s C - C c o s B)$
$s i n (C - A) = k (c c o s A - Q c o s C)$
LHS
$Q s i n (B - C) + b s i n (C - A) + c s i n (A - B)$
$\Rightarrow Q k (b c o s C - c c o s B) + b k (a c o s C - c c o s A)$
$+ C k (a c o s B - b c o s A)$
$\Rightarrow k (b c c o s A) + k (a c c o s B - a c c o s B)$
$+ k (a b c o s C - a b c o s C)$
$= 0$

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