Answer :
Compute the characteristic polynomial of [tex]A[/tex].
[tex]p(\lambda) = \det(A - \lambda I) \\\\ ~~~~~~~~ = \begin{vmatrix} \beta - \lambda & 0 & 1 \\ 2 & 1 - \lambda & -2 \\ 3 & 1 & -2-\lambda \end{vmatrix} \\\\ ~~~~~~~~ = (\beta-\lambda)\begin{vmatrix}1-\lambda&-2\\1&-2-\lambda\end{vmatrix} + \begin{vmatrix}2&1-\lambda \\ 3 & 1 \end{vmatrix} \\\\ ~~~~~~~~ = -\lambda^3 + (\beta-1) \lambda^2 + (\beta + 3) \lambda - 1[/tex]
By the Cayley-Hamilton theorem,
[tex]p(A) = -A^3 + (\beta - 1) A^2 + (\beta + 3) A - I = 0[/tex]
Note that
[tex]A^7 - (\beta - 1) A^6 - \beta A^5 \\\\ ~~~~~~~~ = -(-A^3 + (\beta - 1) A^2 + \beta A) A^4 \\\\ ~~~~~~~~ = -(\underbrace{-A^3 + (\beta - 1) A^2 + (\beta + 3) A - I}_{p(A)=0} - 3A + I) A^4 \\\\ ~~~~~~~~ = (3A - I) A^4[/tex]
Recall that [tex]\det(AB) = \det(A) \det(B)[/tex]. Note that [tex]A[/tex] is not singular, and hence [tex]A^n[/tex] is not singular for any natural [tex]n[/tex].
[tex]\det(A) = \begin{vmatrix} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{vmatrix} \\\\ ~~~~~~~~ = \beta \begin{vmatrix}1&-2\\1&-2\end{vmatrix} + \begin{vmatrix}2&1\\3&1\end{vmatrix} \\\\ ~~~~~~~~ = -1 \neq 0[/tex]
Then [tex]3A-I[/tex] must be singular. So
[tex]\det(3A-I) = \begin{vmatrix} 3\beta - 1 & 0 & 3 \\ 6 & 2 & -6 \\ 9 & 3 & -7 \end{vmatrix} \\\\ ~~~~~~~~ = (3\beta-1) \begin{vmatrix}2&-6\\3&-7\end{vmatrix} + 3 \begin{vmatrix}6&2\\9&3\end{vmatrix} \\\\ ~~~~~~~~ = 4 (3\beta-1)[/tex]
[tex]4(3\beta-1) = 0 \implies \beta = \dfrac13 \implies 9\beta = \boxed{3}[/tex]