Let  $\alpha ,\beta ,\gamma \in R$  such that  $\cos (\alpha -\beta )+\cos (\beta -\gamma )+\cos (\gamma -\alpha )=-\frac{3}{2}$. 

Then match Column – I with  Column – II 

	Column – I		Column – II
A	$\sum \sin (\alpha +\beta )=\sum \cos (\alpha +\beta )=$	p	0
B	$\sum \sin 3\alpha =$	q	$3\sin (\alpha +\beta +\gamma )$
C	$\sum \cos 3\alpha =$	r	$3\cos (\alpha +\beta +\gamma )$
D	If  $\theta \in R$ , then  $\frac{{\displaystyle \sum {\cos }^3(\theta +\alpha )}}{\Pi \cos (\theta +\alpha )}=$	s	3

 $\cos (\alpha -\beta )+\cos (\beta -\gamma )+\cos (\gamma -\alpha )=-3/2$
 $3+2(\cos \alpha \cos \beta +\cos \beta \cos \gamma +\cos \gamma \cos \alpha +\sin \alpha \sin \beta +\sin \beta \sin \gamma +\sin \gamma \sin \alpha )=0$
 ${(\cos \alpha +\cos \beta +\cos \gamma )}^2+{(\sin \alpha +\sin \beta +\sin \gamma )}^2=0$
 $\Rightarrow \cos \alpha +\cos \beta +\cos \gamma =0$                    ……………… (1)
 $\sin \alpha +\sin \beta +\sin \gamma =0$                               ………………. (2)
$\Rightarrow bc+ca+ab=0\Rightarrow cis(\alpha +\beta )+cis(\beta +\gamma )+cis(\gamma +\alpha )=0$$a=cis\alpha ,b=cis\beta ,c=cis\gamma ,\Rightarrow a+b+c=0$  and  $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$  (by (1) & (2))
 $\Rightarrow bc+ca+ab=0\Rightarrow cis(\alpha +\beta )+cis(\beta +\gamma )+cis(\gamma +\alpha )=0$
And  $\sin (\alpha +\beta )+\sin (\beta +\gamma )+\sin (\gamma +\alpha )=0$
   $\therefore \sum \sin (\alpha +\beta )=\sum \cos (\alpha +\beta )=0$     $\therefore \left(A\right)\Rightarrow P$
We have  $a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\Rightarrow \sum cis3\alpha =3cis(\alpha +\beta +\gamma )$
By compare imaginary part  $\sum \cos 3\alpha =3\cos (\alpha +\beta +\gamma )\Rightarrow \therefore B\to qandc\to r$
 $\theta \in R,\frac{{\displaystyle \sum {\cos }^3(\theta +\pi )}}{\pi \cos (\theta +\alpha )}=\frac{3\pi \cos (\theta +\alpha )}{\pi \cos (\theta +\alpha )}=3$