The maximum value

$\Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1+\cos \theta & 1 & 1\end{array}\right|$

$=\left|\begin{array}{ccc}1 & 1 & 1 \\ 0 & \sin \theta & 0 \\ \cos \theta & 0 & 0\end{array}\right|$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$ ]

$=-\sin \theta \cos \theta$

$=-\frac{\sin 2 \theta}{2}$

Now, Maximum and minimum value of $\sin \theta$ is 1 and $-1$.

So, the maximum value of $-\sin \theta$ is 1 .

So, the maximum value of $-\sin 2 \theta$ is 1 .

Therefore, the maximum value of $-\frac{\sin 2 \theta}{2}$ is $\frac{1}{2}$.

Hence, the correct option is (a).