In $△ A B C, a sin (B - C) + b sin (C - A) + c sin (A - B) =$

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 $0$
$a sin (B - c) + b sin (c - A) + c sin (A - B)$
Sin Rule $\Rightarrow \frac{sin A}{a} = sin B - sin C - k$
So write $sin A = a K, sin B = b k, sin C = c k$
As we know $sin (A - B) = sin A cos B - cos A sin B$
$\begin{matrix} Put value of sin A and sin B sin (A - B) = akcos B - b K cos A = K (2 cos B - b cos A) \end{matrix}$
similarity for $sin (c - A) = k (c cos A - a cos C)$
$sin (B - C) = k (b cos C - c cos B)$
$\begin{matrix} = a k (b cos c - c cos B) + b k (c cos A - a cos c) + c k (a cos B - b cos β) = k (b c cos A - b c (θ s A) + k (a c cos B - a c cos B) + k (a b cos c - a b cos c \end{matrix}$
$= 0 + 0 + 0 = 0$
$a sin (B - C) + bsin (C - A) + c sin (A - B) = 0$
A option is correct.

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|\begin{matrix} - a (b^{2} + c^{2} - a^{2}) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & - b (c^{2} + a^{2} - b^{2}) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & - c (a^{2} + b^{2} - c^{2}) \end{matrix}| = a b c {(a^{2} + b^{2} + c^{2})}^{3}

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