Explanation

(b)
Given that, $\left| \begin{matrix}
x & - 6 & - 1 \\
2 & - 3x & x - 3 \\
- 3 & 2x & x + 2
\end{matrix} \right|$=0
$\Rightarrow \ \left| \begin{matrix}
x & - 6 & - 1 \\
5 & - 5x & - 5 \\
- 3 & 2x & x + 2
\end{matrix} \right| = 0$ $(R_{2} \rightarrow R_{2} - R_{3})$
$\Rightarrow 5\left| \begin{matrix}
x & - 6 & - 1 \\
1 & - x & - 1 \\
- 3 & 2x & x + 2
\end{matrix} \right| = 0$
$\Rightarrow x\left( - x^{2} - 2x + 2x \right) - 1( - 6x - 12 + 2x) - 3(6 - x) = 0$
$\Rightarrow \ - x^{3} + 7x - 6 = 0$
$\Rightarrow x^{3} - 7x + 6 = 0$
$\Rightarrow (x - 1)(x - 2)(x + 3) = 0$
$\Rightarrow x = 1,\ 2,\ - 3$
$\therefore$ Option (b) is correct.