Let $\omega $ be a solution of ${x^3} - 1 = 0$ with $\operatorname{Im} \left( \omega \right) > 0$. If $a = 2$ with b and c satisfying
\[\left[ {\begin{array}{*{20}{c}}
  a&b&c
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&9&7 \\
  8&2&7 \\
  7&3&7
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0&0
\end{array}} \right]......\left( E \right)\]
Then the value of $\dfrac{3}{{{\omega ^a}}} + \dfrac{1}{{{\omega ^b}}} + \dfrac{3}{{{\omega ^c}}}$ is :
(A) $ - 2$
(B) $2$
(C) $3$
(D) $ - 3$

Hint: The multiplication of two matrices is possible if the no. of columns in matrix A is equal to the no. of rows in matrix B. Here we multiplied the two given matrix and form the equations by comparing the values of both sides.

Complete step-by-step answer:
Since, \[a,b\] and $c$ be three real numbers satisfies
\[\left[ {\begin{array}{*{20}{c}}
  a&b&c
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&9&7 \\
  8&2&7 \\
  7&3&7
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&0&0
\end{array}} \right]\]
So, we get the equations
$
  a + 8b + 7c = 0 \\
  9a + 2b + 3c = 0 \\
  7a + 7b + 7c = 0 \Rightarrow a + b + c = 0 \\
 $
Since $a = 2$, so the equations become
$
  2 + 8b + 7c = 0 \Rightarrow 8b + 7c = - 2....(1) \\
  9\left( 2 \right) + 2b + 3c = 0 \Rightarrow 2b + 3c = - 18....(2) \\
  2 + b + c = 0 \Rightarrow b + c = - 2....(3) \\
 $
Multiply equation (3) by $7$ and subtract it from (1), we get
$8b + 7c - \left( {7b + 7c} \right) = - 2 - \left( { - 14} \right)$
$
   \Rightarrow 8b + 7c - 7b - 7c = - 2 + 14 \\
   \Rightarrow b = 12 \\
 $
 Substitute the value of $b$ in equation (3), we get
$
  12 + c = - 2 \\
   \Rightarrow c = - 2 - 12 \\
   \Rightarrow c = - 14 \\
 $
So, we have $a = 2,b = 12,c = - 14$
Now, $\dfrac{3}{{{\omega ^a}}} + \dfrac{1}{{{\omega ^b}}} + \dfrac{3}{{{\omega ^c}}}$$ = \dfrac{3}{{{\omega ^2}}} + \dfrac{1}{{{\omega ^{12}}}} + \dfrac{3}{{{\omega ^{ - 14}}}}$
\[ = \dfrac{3}{{{\omega ^2}}} + \dfrac{1}{{{\omega ^{12}}}} + 3{\omega ^{14}}\]
Convert the power of $\omega $ in terms of ${\omega ^3}$, because ${\omega ^3} = 1$:-
\[ = \dfrac{{3\omega }}{{{\omega ^3}}} + \dfrac{1}{{{{\left( {{\omega ^3}} \right)}^4}}} + 3{\left( {{\omega ^3}} \right)^4} \cdot {\omega ^2}\]
Put ${\omega ^3} = 1$ and simplify the above,
\[ = \dfrac{{3\omega }}{1} + \dfrac{1}{{{{\left( 1 \right)}^4}}} + 3{\left( 1 \right)^4} \cdot {\omega ^2}\]
\[ = 3\omega + 1 + 3{\omega ^2}\]
$ = 3\left( {\omega + {\omega ^2}} \right) + 1$
Since $1 + \omega + {\omega ^2} = 0$$ \Rightarrow \omega + {\omega ^2} = - 1$
$ = 3\left( { - 1} \right) + 1$
$
   = - 3 + 1 \\
   = - 2 \\
 $
So, $\dfrac{3}{{{\omega ^a}}} + \dfrac{1}{{{\omega ^b}}} + \dfrac{3}{{{\omega ^c}}}$$ = - 2$

Hence, option (A) is the correct answer.

Note: Remember the formulae regarding $\omega $ i.e., ${\omega ^3} = 1,1 + \omega + {\omega ^2} = 0$ to solve these types of problems. Also, it is important to reduce the high powers of $\omega $into small powers of $\omega $, as we have done in this question.