Paragraph:
Let \(a\), b and \(c\) be three real numbers satisfying \(\left[\begin{array}{lll}a & b & c\end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]\)
Question:
Let \(\omega\) be a solution of \(x^3-1=0\) with \(\operatorname{Im}(\omega)>0\). If \(a=2\), with \(b\) and \(c\) satisfying Eq. (E), then the value of \(\frac{3}{\omega^a}+\frac{1}{\omega^b}+\frac{3}{\omega^c}\)

\(
\begin{aligned}
& \text {Given, }[a b c]_{1 \times 3}\left[\begin{array}{lll}
1 & 9 & 7 \\
8 & 2 & 7 \\
7 & 3 & 7
\end{array}\right]_{3 \times 3}=\left[\begin{array}{lll}
0 & 0 & 0
\end{array}\right] \\
& \Rightarrow \left[\begin{array}{c}
a+8 b+7 c \\
9 a+2 b+3 c \\
7 a+7 b+7 c
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] \\
& \Rightarrow a+8 b+7 c=0 \\
& \Rightarrow 9 a+2 b+3 c=0 \\
& \Rightarrow a+b+c=0\end{aligned}
\)
On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get
\(
7 a+c=0
\)
Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get
\(6 a-b=0\)
\(\therefore b=6 a \text { and } c=-7 a\)
If \(a=2, b=12\) and \(c=-14\)
\(\therefore \frac{3}{\omega^a}+\frac{1}{\omega^b}+\frac{3}{\omega^c} \)
\( \Rightarrow \frac{3}{\omega^2}+\frac{1}{\omega^{12}}+\frac{3}{\omega^{-14}}=\frac{3}{\omega^2}+1+3 \omega^2 \)
\( =3 \omega+1+3 \omega^2 \)
\( =1+3\left(\omega+\omega^2\right) \)
\( =1-3=-2\)