a
$A=\left[\begin{array}{ccc}\alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta\end{array}\right]$
$R _{3} \rightarrow R _{3}+ R _{1}$
$| A |=|\alpha+\beta+\gamma|\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ 1 & 1 & 1\end{array}\right|$
$\Rightarrow| A |=(\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$
$\because|\operatorname{adj} A |=| A |^{ n -1}$
$|\operatorname{adj}(\operatorname{adj} A )|=| A |^{( n -1)^{2}}$
$|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A )))|=| A |^{( n -1)^{4}}=| A |^{2^{4}}=| A |^{16}$
$\therefore(\alpha+\beta+\gamma)^{16}=2^{32} \cdot 3^{16}$
$(\alpha+\beta+\gamma)^{16}=\left(2^{2} \cdot 3\right)^{16}=(12)^{16}$
$\alpha+\beta+\gamma=12$
$\because \alpha, \beta, \gamma \in N$
$(\alpha-1)+(\beta-1)+(\gamma-1)=9$
number all tuples $(\alpha, \beta, \gamma)={ }^{11} C _{2}=55$
$1$ case for $\alpha=\beta=\gamma$
$12$ case when any two of these are equal
So, No. of distinct tuples $(\alpha, \beta, \gamma)$
$=55-13=42$