$${A^7} - (\beta - 1){A^6} - \beta {A^5}$$ is a singular matrix. So determinant of this matrix equal to zero.

$$\therefore$$ $$|{A^7} - (\beta - 1){A^6} - \beta {A^5}| = 0$$

$$ \Rightarrow |{A^5}({A^2} - (\beta - 1)A - \beta I)| = 0$$

$$ \Rightarrow |{A^5}||({A^2} - \beta A + A - \beta I)| = 0$$

$$ \Rightarrow |A{|^5}|A(A + I) - \beta (A + I)| = 0$$

$$ \Rightarrow |A{|^5}|(A - \beta I)(A + I)| = 0$$

$$ \Rightarrow |A{|^5}|A - \beta I||A + I| = 0$$

Now given,

$$A = \left[ {\matrix{ \beta & 0 & 1 \cr 2 & 1 & { - 2} \cr 3 & 1 & { - 2} \cr } } \right]$$

$$\therefore$$ $$|A| = 2 - 3 = - 1$$

$$|A + I| = \left| {\matrix{ {\beta + 1} & 0 & 1 \cr 2 & 2 & { - 2} \cr 3 & 1 & { - 1} \cr } } \right|$$

$$ = (\beta + 1)( - 2 + 2) + 1(2 - 6)$$

$$ = - 4$$

$$\therefore$$ We get $$|A| \ne 0$$ and $$|A + I| \ne 0$$

$$\therefore$$ $$|A{|^5}|A - \beta I||A + I| = 0$$ is possible only when $$|A - \beta I| = 0$$

$$\therefore$$ $$|A - \beta I| = \left| {\matrix{ 0 & 0 & 1 \cr 2 & {1 - \beta } & { - 2} \cr 3 & 1 & { - 2 - \beta } \cr } } \right|$$

$$ = 2 - 3 - 3\beta $$

$$\therefore$$ $$2 - 3 + 3\beta = 0$$

$$ \Rightarrow 3\beta = 1$$

$$ \Rightarrow 9\beta = 3$$