If $a+b+c=\pi$, then the value of $\left|\begin{array}{ccc}\sin (a+b+c) & \sin (a+c) & \cos b \\ -\sin b & 0 & \tan a \\ \cos (a+c) & \tan (b+c) & 0\end{array}\right|$ is:

0

The correct answer is Option (1) - 0

$Δ=\left|\begin{array}{ccc}\sin (a+b+c) & \sin (a+c) & \cos b \\ -\sin b & 0 & \tan a \\ \cos (a+c) & \tan (b+c) & 0\end{array}\right|$

$Δ=\begin{vmatrix}\sin π&\sin(π-b)&\cos b\\=\sin b&0&\tan a\\\cos(π-b)&\tan(π-a)&0\end{vmatrix}$

$Δ=\begin{vmatrix}0&\sin b&\cos b\\-\sin b&0&\tan a\\-\cos b&-\tan b&0\end{vmatrix}$

here it is a skew symmetric matrix

so $Δ=0$