Answer:
Option A
Explanation:
Given ,[a b c]_{1 \times 3} \begin{bmatrix}1 & 9 &7 \\8 & 2&7 \\7&3&7 \end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}
\Rightarrow \begin{bmatrix}a+8b+7c \\9a+2b+3c \\7a+7b+7c\end{bmatrix}=\begin{bmatrix}0 & 0 &0 \end{bmatrix}
\Rightarrow a+8b+7c=0....(i)
\Rightarrow 9a+2b+3c=0 .....(ii)
\Rightarrow a+b+c=0 .....(iii)
On multiplying Eq.(iii) bt=y 2 , then substract from Eq.(ii) we get
7a+c=0...(iv)
Again multiplying Eq,(iii) by 3 , then substract from Eq.(ii) , we get
6a-b=0....(v)
\therefore b=6a and c=-7a
if a=2,b=12 and c=-14
\therefore \frac{3}{ \omega^{a}}+\frac{1}{\omega^{b}}+\frac{3}{\omega^{c}}
\Rightarrow \frac{3}{\omega^{2}}+\frac{1}{\omega^{12}}+\frac{3}{\omega^{-14}}
= \frac{3}{ \omega^{2}}+1+3 \omega^{2}
=3 \omega+1+ 3 \omega^{2}
=1+3(\omega+\omega^{2})
=1-3=-2