# In a triangle ABC $\begin{vmatrix} a^2& b\sin A & C\sin A\\b\sin A&1&\cos(B-C)\\C\sin A&\cos (B-C) & 1\end{vmatrix}$ equals

$\begin{array}{1 1}(A)\;\sin A-\sin B\sin C&(B)\;abc\\(C)\;1&(D)\;0\end{array}$

Using the sine formula
$\large\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{C}{\sin C}=\frac{1}{k}$(say)
$\sin A=ak,\sin B=bk,\sin C=ck$
Given
Determinant=$\begin{vmatrix} a^2& b\sin A & C\sin A\\b\sin A&1&\cos(B-C)\\C\sin A&\cos (B-C) & 1\end{vmatrix}$
$\Rightarrow \begin{vmatrix} 1& \sin B & \sin C\\\sin B&1&\cos(B-C)\\\sin C&\cos (B-C) & 1\end{vmatrix}$
$\Rightarrow \begin{vmatrix} 1& \sin (A+C) & \sin (A+B)\\\sin (A+C)&1&\cos(B-C)\\\sin (A+B)&\cos (B-C) & 1\end{vmatrix}$
$\Rightarrow a^2\begin{vmatrix}\sin A&\cos A &0\\\cos C&\sin C&0\\\cos B& \sin B& 0\end{vmatrix}\begin{vmatrix}\sin A&\cos A &0\\\cos C&\sin C&0\\\cos B& \sin B& 0\end{vmatrix}=0$
Hence (D) is the correct answer.