Consider a matrix A = [ cases α & β & γ α ^ 2 & β ^ 2 & γ ^ 2 β + γ...

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

${R}_{3} \to {R}_{3} + {R}_{1}$

A = \left[$\begin{cases}\alpha & \beta & \gamma \\ {\alpha}^{2} & {\beta}^{2} & {\gamma}^{2} \\ \alpha + \beta + \gamma & \alpha + \beta + \gamma & \alpha + \beta + \gamma\end{cases}$\right]$</p><p>$⇒ \left|A\right| = \left|\alpha + \beta + \gamma\right|\left|$\begin{cases}\alpha & \beta & \gamma \\ {\alpha}^{2} & {\beta}^{2} & {\gamma}^{2} \\ 1 & 1 & 1\end{cases}$\right|$</p><p>$⇒ \left|A\right| = \left(\alpha + \beta + \gamma\right)\left(\alpha - \beta\right)\left(\beta - \gamma\right)\left(\gamma - \alpha\right)$</p><p>We know \left|adj A\right|$ = {\left|A\right|}^{n - 1}

⇒ \left|adj \left(adj A\right)\right| = {\left|A\right|}^{{\left(n - 1\right)}^{2}}$</p><p>$⇒ \left|adj \left(adj \left(adj \left(adj A\right)\right)\right)\right| = {\left|A\right|}^{{\left(n - 1\right)}^{4}}

Here \left|adj \left(adj \left(adj \left(adj A\right)\right)\right)\right| = {\left|A\right|}^{{2}^{4}} = {\left|A\right|}^{16}$</p><p>Given$\dfrac{\left|\left(adj ( adj \left(adj\left(adj A\right)\right)\right)\right|}{{\left(\alpha - \beta\right)}^{16}{\left(\beta - \gamma\right)}^{16}{\left(\gamma - \alpha\right)}^{16}} = {2}^{32} \times {3}^{16}$</p><p>$⇒ \dfrac{{\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}$</p><p>$∴ {\left(\alpha + \beta + \gamma\right)}^{16} = {2}^{32} \cdot {3}^{16}$</p><p>$⇒ {\left(\alpha + \beta + \gamma\right)}^{16} = {\left(12\right)}^{16}$</p><p>$⇒ \alpha + \beta + \gamma = 12$</p></blockquote><blockquote><p>$∵ \alpha, \beta, \gamma \in N$</p><p>\left(\alpha - 1\right)$ + \left(\beta - 1\right)$+ \left(\gamma - 1\right)$ = 9$</p><p>Possible number all tuples \left(\alpha, \beta, \gamma\right) will be$C2 11 = 55$</p><p>$1$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 $distinct tuples \left(\alpha, \beta, \gamma\right)</p><p>$= 55 - 13 = 42$</p></blockquote>$