Answer: Option B
:
B
cos$\alpha$+cos$\beta$+cos$\gamma$ = 0............(i)
sin$\alpha$+sin$\beta$+sin$\gamma$ =0...........(ii)
Let a=cos$\alpha$+isin$\alpha$, b=cos$\beta$+isin$\beta$
c=cos$\gamma$+isin$\gamma$
$\Rightarrow$ a+b+c=0[By (i) and (ii)]....(iii)
$\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$
=cos$\alpha$-isin$\alpha$+cos$\beta$-isin$\beta$+cos$\gamma$-isin$\gamma$
$\Rightarrow$ ab+bc+ca=0 ......(iv) [by (i) and (ii)]
Squaring both sides of equation (iii),
we get $a^2$+$b^2$+$c^2$+2ab+2bc+2ca=0
or $a^2$+$b^2$+$c^2$=0 [by (iv)]
$[cos\alpha +isin\alpha {]}^2$ + $[cos\beta +isin\beta {]}^2$ + $[cos\gamma +isin\gamma {]}^2$ = 0
$\Rightarrow$ (cos2$\alpha$+cos2$\beta$+cos2$\gamma$) + i(sin2$\alpha$+sin2$\beta$+sin2$\gamma$)=0
Separation of real and imaginary part,
$\Rightarrow$ cos2$\alpha$+cos2$\beta$+cos2$\gamma$=0 .........(v)
And sin2$\alpha$+sin2$\beta$+sin2$\gamma$ =0 .........(iv)
$\Rightarrow$ 1-2$si{n}^2$$\alpha$+1-2$si{n}^2$$\beta$+1-2$si{n}^2$$\gamma$=0 [By eq.(v)]
Therefore, $si{n}^2$$\alpha$+$si{n}^2$$\beta$+$si{n}^2$$\gamma$= $\frac{3}{2}$