Mark the correct alternative in each of the following:

Solution:

Using sine rule, we have

$\sum a^{2}(\sin B-\sin C)$

$=a^{2}\left(\frac{b}{k}-\frac{c}{k}\right)+b^{2}\left(\frac{c}{k}-\frac{a}{k}\right)+c^{2}\left(\frac{a}{k}-\frac{b}{k}\right)$

 

$=\frac{1}{k}\left(a^{2} b-a^{2} c+b^{2} c-b^{2} a+c^{2} a-c^{2} b\right)$

This expression cannot be simplified to match with any of the given options.

However, if the quesion is "In any $\triangle \mathrm{ABC}, \sum a^{2}\left(\sin ^{2} B-\sin ^{2} C\right)=$ ", then the solution is as follows.

Using sine rule, we have

$\sum a^{2}\left(\sin ^{2} B-\sin ^{2} C\right)$

$=a^{2}\left(\frac{b^{2}}{k^{2}}-\frac{c^{2}}{k^{2}}\right)+b^{2}\left(\frac{c^{2}}{k^{2}}-\frac{a^{2}}{k^{2}}\right)+c^{2}\left(\frac{a^{2}}{k^{2}}-\frac{b^{2}}{k^{2}}\right)$

 

$=\frac{1}{k^{2}}\left(a^{2} b^{2}-a^{2} c^{2}+b^{2} c^{2}-b^{2} a^{2}+c^{2} a^{2}-c^{2} b^{2}\right)$

$=\frac{1}{k^{2}} \times 0$

 

$=0$

Hence, the correct answer is option (d).

Disclaimer: The question given in the book in incorrect or there is some printing mistake in the question.