Given $\mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{3\mathrm{\pi }}{2} ...\left(1\right)$

Now 

$\cos 2\mathrm{A}+\cos 2\mathrm{B}+\cos 2\mathrm{C}$

$=2\cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})+\cos 2\mathrm{C}$

$=2\cos \left(\frac{3\mathrm{\pi }}{2}-\mathrm{C}\right)\cos (\mathrm{A}-\mathrm{B})+\cos 2\mathrm{C}$
$=-2\mathrm{sinC}\times \cos (\mathrm{A}-\mathrm{B})+1-2{\sin }^2\mathrm{C}$

$=-2\mathrm{sinC} \left[\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B})\right]+1$
$=1-4\mathrm{sinAsinBsin} \mathrm{C}$

Now putting the value of $\cos 2\mathrm{A}+\cos 2\mathrm{B}+\cos 2\mathrm{C}$ in $4\sin A\sin B\sin C+\cos 2A+\cos 2B+\cos 2C$ we get,

$4\sin A\sin B\sin C+\cos 2A+\cos 2B+\cos 2C=1$

Now solving options we get,

$-\sin \left(A+B+C\right)=-\sin \left(\frac{3\mathrm{\pi }}{2}\right)=-\left(-1\right)=1$