# If $A,B$ and $C$ are the angles of a $\Delta$le then $\begin{vmatrix}\sin 2A&\sin C&\sin B\\\sin C&\sin 2B&\sin A\\\sin B&\sin A&\sin 2C\end{vmatrix}$ = $\lambda\sin A\sin B\sin C$, then $\lambda$ is

$(A)\;1\qquad(B)\;0\qquad(C)\;2\qquad(D)\;3$

LHS $\begin{vmatrix}\sin 2A&\sin C&\sin B\\\sin C&\sin 2B&\sin A\\\sin B&\sin A&\sin 2C\end{vmatrix}$ from sine rule we have
$\Rightarrow \begin{vmatrix}2ka\cos A&kc&kb\\kb&2kb\cos B&ka\\kb&ka&2kc\cos C\end{vmatrix}$
$\Rightarrow k^3\begin{vmatrix}2a\cos A&c&b\\c&2b\cos B&a\\b&a&2c\cos C\end{vmatrix}$
$\Rightarrow k^3\begin{vmatrix}a\cos A+a\cos A&a\cos B+b\cos A&a\cos C+c\cos A\\a\cos B+b\cos A&b\cos B+b\cos B&b\cos C+c\cos B\\a\cos C+c\cos A&b\cos C+c\cos B&c\cos C+c\cos C\end{vmatrix}$
$\Rightarrow k^3\begin{vmatrix}\cos A&a&0\\\cos B&b&0\\\cos C&c&0\end{vmatrix}\times \begin{vmatrix}a&\cos A&0\\b&\cos B&0\\c&\cos C&0\end{vmatrix}$
$\Rightarrow 0\times 0=0$
$\Rightarrow \lambda=0$
Hence (b) is the correct answer.