Match List-I with List-II and select the correct answer using the code given below the list.

	COLUMN-I		COLUMN-II
A)	Let  $\alpha ,\beta ,\gamma ,\delta$  be the roots of the equation  $x^4-x^3+2x^2+x+1=0$ , then the value of  $({\alpha }^3+1)({\beta }^3+1)({\gamma }^3+1)({\delta }^3+1)$  is	P)	0
B)	The value of $\frac{2^{180}}{45}\prod _{k=1}^{179}\sin k^0$  is	Q)	4
C)	If complex numbers  $z_1,z_3$ and  $z_3$ are the vertices of an isosceles right angled triangle, right angled at the vertex  $z_2$, then  ${(z_1-z_2)}^2+{(z_3-z_2)}^2$ equals	R)	8
D)	Let  $Z=\frac{1}{2}(\sqrt{3}-i)$. If the least positive integer value of ‘n’ such that ${(z^{101}+i^{109})}^{106}=z^n$  is ‘k’ then $\frac{2}{5}k$  is	S)	16

A)  $({\alpha }^3+1)({\beta }^3+1)({\gamma }^3+1)({\delta }^3+1)$
 $$\begin{gathered} T1(\alpha +1)(\alpha +\omega )(\alpha +{\omega }^2) \\ =f(-1)f(-\omega )f(-{\omega }^2) \\ =\left(4\right)(2{\omega }^2+2)(2\omega +2) \\ =\left(4\right)\left(4\right)=16 \end{gathered}$$
B)  $\prod _{k=1}^{179}\sin k^{\circ }=\sin 1^{\circ }\sin 2^{\circ }-------\sin {179}^{\circ }{z}^{180}=1$
 $$\begin{gathered} z^{180}-1=(z-1)(z-\alpha )-----(z-{\alpha }^{179}) \\ \frac{z^{180}-1}{z-1}=(z-\alpha )(z-{\alpha }^2)------(z-{\alpha }^{179}) \\ \Rightarrow \left(180\right)={\left(12\right)}^{179}\sin 1^{\circ }\sin 2^{\circ }-----\sin {179}^{\circ } \\ \sin 1^{\circ }\sin 2^{\circ }------\sin {179}^{\circ }=\frac{180}{2^{179}} \\ \frac{2^{180}}{45}\sin 1^{\circ }\sin 2^{\circ }------\sin {179}^{\circ }=2^3=8 \end{gathered}$$
C. we know that  $\frac{z_3-z_2}{z_1-z_2}=\frac{|z_3-z_2|}{|z_1-z_2|}.e^{\frac{i\pi }{2}}$
 $\Rightarrow z_3-z_2=i(z_1-z_2)\Rightarrow {(z_3-z_2)}^2+{(z_1+z_2)}^2=0$
D.    $z=\frac{-1}{2}i(1+i\sqrt{3})=i{\omega }^2$                     $z^{101}=i\omega$
 ${(z^{101}+i^{100})}^{106}=-{\left(i{\omega }^2\right)}^{106}=-{\omega }^2$
   $\therefore -{\omega }^2={\left(i{\omega }^2\right)}^{2n}=i^{\pi }{\omega }^{2n} {\omega }^{2\pi -2}{i}^{\pi }=-1$
This is possible only when N + 4r + 2 and 2x -2 is multiple of 3 i.e., 2 (4r + 2) – 2 is multiply of 3
i.e., 8r + 2 is multiple of  $3\Rightarrow r=2\therefore x=10\therefore \frac{2}{5}k=4$