Consider a matrix $\mathrm{A}=\left[\begin{array}{ccc}\mathrm{\alpha }& \mathrm{\beta }& \mathrm{\gamma }\\ {\mathrm{\alpha }}^2& {\mathrm{\beta }}^2& {\mathrm{\gamma }}^2\\ \mathrm{\beta }+\mathrm{\gamma }& \mathrm{\gamma }+\mathrm{\alpha }& \mathrm{\alpha }+\mathrm{\beta }\end{array}\right],$ where $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }$ are three distinct natural numbers.

If $\frac{\det (\mathrm{adj}(\mathrm{adj}(\mathrm{adj}(\mathrm{adj}\mathrm{A}\left)\right)\left)\right)}{(\mathrm{\alpha }-\mathrm{\beta }{)}^{16}(\mathrm{\beta }-\mathrm{\gamma }{)}^{16}(\mathrm{\gamma }-\mathrm{\alpha }{)}^{16}}=2^{32}\times 3^{16},$ then the number of such 3 – tuples ($\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }$) is _______ .

$\begin{array}{l}\mathrm{A}=\left[\begin{array}{ccc}\mathrm{\alpha }& \mathrm{\beta }& \mathrm{\gamma }\\ {\mathrm{\alpha }}^2& {\mathrm{\beta }}^2& {\mathrm{\gamma }}^2\\ \mathrm{\beta }+\mathrm{\gamma }& \mathrm{\gamma }+\mathrm{\alpha }& \mathrm{\alpha }+\mathrm{\beta }\end{array}\right]\\ {\mathrm{R}}_3\to {\mathrm{R}}_3+{\mathrm{R}}_1\\ \Rightarrow \left|\mathrm{A}\right|=|\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }|\left|\begin{array}{lcc}\mathrm{\alpha }& \mathrm{\beta }& \mathrm{\gamma }\\ {\mathrm{\alpha }}^2& {\mathrm{\beta }}^2& {\mathrm{\gamma }}^2\\ 1& 1& 1\end{array}\right|\\ \Rightarrow \left|\mathrm{A}\right|=(\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma })(\mathrm{\alpha }-\mathrm{\beta })(\mathrm{\beta }-\mathrm{\gamma })(\mathrm{\gamma }-\mathrm{\alpha })\\ \because |\mathrm{adj}\mathrm{A}|=|\mathrm{A}{|}^{\mathrm{n}-1}\\ |\mathrm{adj}(\mathrm{adj}\mathrm{A}\left)\right|=|\mathrm{A}{|}^{(\mathrm{n}-1{)}^2}\\ |\mathrm{adj}(\mathrm{adj}(\mathrm{adj}(\mathrm{adj}\mathrm{A}\left)\right)\left)\right|=|\mathrm{A}{|}^{(\mathrm{n}-1{)}^4}=|\mathrm{A}{|}^{2^4}=|\mathrm{A}{|}^{16}\\ \therefore (\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }{)}^{16}=2^{32}\cdot 3^{16}\\ \Rightarrow (\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }{)}^{16}={\left(2^2\cdot 3\right)}^{16}=(12{)}^{16}\\ \Rightarrow \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=12\\ \because \mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }\in \mathrm{N}\\ (\mathrm{\alpha }-1)+(\mathrm{\beta }-1)+(\mathrm{\gamma }-1)=9\end{array}$

number all tuples $(\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }){=}^{11}{\mathrm{C}}_2=55$

1 case for $\mathrm{\alpha }=\mathrm{\beta }=\mathrm{\gamma }$

& 12 case when any two of these are equal

So, No. of distinct tuples $\left(\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }\right)=55-13=42$