Consider a matrix A = α β γ α 2 β 2 γ 2 β + γ γ + α α…

Given $A=\left[\begin{array}{ccc}\alpha & \beta & \gamma \\ {\alpha }^2& {\beta }^2& {\gamma }^2\\ \beta +\gamma & \gamma +\alpha & \alpha +\beta \end{array}\right]$

$R_3\to R_3+R_1$

$A=\left[\begin{array}{ccc}\alpha & \beta & \gamma \\ {\alpha }^2& {\beta }^2& {\gamma }^2\\ \alpha +\beta +\gamma & \alpha +\beta +\gamma & \alpha +\beta +\gamma \end{array}\right]$

$\Rightarrow \left|A\right|=\left|\alpha +\beta +\gamma \right|\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ {\alpha }^2& {\beta }^2& {\gamma }^2\\ 1& 1& 1\end{array}\right|$

$\Rightarrow \left|A\right|=\left(\alpha +\beta +\gamma \right)\left(\alpha -\beta \right)\left(\beta -\gamma \right)\left(\gamma -\alpha \right)$

We know $\left|adjA\right|={\left|A\right|}^{n-1}$

$\Rightarrow \left|adj\left(adjA\right)\right|={\left|A\right|}^{{\left(n-1\right)}^2}$

$\Rightarrow \left|adj\left(adj\left(adj\left(adjA\right)\right)\right)\right|={\left|A\right|}^{{\left(n-1\right)}^4}$

Here $\left|adj\left(adj\left(adj\left(adjA\right)\right)\right)\right|={\left|A\right|}^{2^4}={\left|A\right|}^{16}$

Given $\frac{\left|\left(adj(adj\left(\mathrm{adj}\left(adjA\right)\right)\right)\right|}{{\left(\alpha -\beta \right)}^{16}{\left(\beta -\gamma \right)}^{16}{\left(\gamma -\alpha \right)}^{16}}=2^{32}\times 3^{16}$

$\Rightarrow \frac{{\left(\alpha +\beta +\gamma \right)}^{16}{\left(\alpha -\beta \right)}^{16}{\left(\beta -\gamma \right)}^{16}{\left(\gamma -\alpha \right)}^{16}}{{\left(\alpha -\beta \right)}^{16}{\left(\beta -\gamma \right)}^{16}{\left(\gamma -\alpha \right)}^{16}}=2^{32}\times 3^{16}$

$\therefore {\left(\alpha +\beta +\gamma \right)}^{16}=2^{32}·3^{16}$

$\Rightarrow {\left(\alpha +\beta +\gamma \right)}^{16}={\left(12\right)}^{16}$

$\Rightarrow \alpha +\beta +\gamma =12$

$\because \alpha , \beta , \gamma \in N$

$\left(\alpha -1\right)+\left(\beta -1\right)+\left(\gamma -1\right)=9$

Possible number all tuples $\left(\alpha ,\beta ,\gamma \right)$ will be ${}^{11}C_2=55$

$1$ case for $\alpha =\beta =\gamma$ and $12$ case when any two of these are equal are also included here but $\alpha \ne \beta \ne \gamma$

Hence, number of distinct tuples $\left(\alpha ,\beta ,\gamma \right)$

$=55-13=42$