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Mathematics Question on matrix transformation

Consider a matrix A=[αβγα2β2γ2β+γγ+αα+β]\begin{array}{l} \text{Consider a matrix}~A=\begin{bmatrix}\alpha & \beta & \gamma \\\\\alpha^2 & \beta^2 & \gamma^2 \\\\\beta+\gamma & \gamma+\alpha & \alpha+\beta \\\\\end{bmatrix}\end{array}

where α,β,γα, β, γ are three distinct natural numbers.
If det(adj(adj(adj(adj A))))(α−β)16(β−γ)16(γ−α)16=232×316,\begin{array}{l}\frac{\text{det(adj(adj(adj(adj A))))}}{\left(\alpha-\beta\right)^{16}\left(\beta-\gamma\right)^{16}\left(\gamma-\alpha\right)^{16}}=2^{32}\times3^{16},\end{array}
then the number of such 3-tuples (α,β,γ)(α, β, γ) is _________.

Answer

det(A)=∣αβγα2β2γ2β+γγ+αα+β∣\begin{array}{l} \text{det}\left(A\right)=\begin{vmatrix}\alpha & \beta & \gamma \\\\\alpha^2 & \beta^2 & \gamma^2 \\\\\beta+\gamma & \gamma+\alpha & \alpha+\beta \\\\\end{vmatrix}\end{array}

R3→R3+R1\begin{array}{l} R_3 \rightarrow R_3 + R_1\end{array}

⇒(α+β+γ)∣αβγα2β2γ2\111∣\begin{array}{l} \Rightarrow \left(\alpha+\beta+\gamma\right)\begin{vmatrix}\alpha & \beta & \gamma \\\\\alpha^2 & \beta^2 & \gamma^2 \\\1 & 1 & 1 \\\\\end{vmatrix}\end{array}

∴det(A)=(α+β+γ)(α–β)(β–γ)(γ–α)\begin{array}{l}\therefore det \left(A\right) = \left(\alpha + \beta + \gamma\right) \left(\alpha – \beta\right) \left(\beta – \gamma\right) \left(\gamma – \alpha\right)\end{array}
Also, det (adj (adj (adj (adj (A)))))
=(det(A))24=(det(A))16\begin{array}{l} =\left(\text{det}\left(A\right)\right)^{2^4}=\left(\text{det}\left(A\right)\right)^{16}\end{array}
∴ (α+β+γ)16(α−β)16(β−γ)16(γ−α)16(α−β)16(β−γ)16(γ−α)16=(4.3)16\begin{array}{l} \therefore\ \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}}=\left(4.3\right)^{16}\end{array}
⇒α+β+γ=12\begin{array}{l}\Rightarrow \alpha + \beta + \gamma = 12 \end{array}
⇒(α,β,γ)distinct natural triplets\begin{array}{l}\Rightarrow \left(\alpha, \beta, \gamma\right) \text{distinct natural triplets}\end{array}
=11C2–1–3C2(4)=55–1–12 =42\begin{array}{l}= ^{11}C_2 – 1 – ^3C_2 \left(4\right) = 55 – 1 – 12\\\ = 42 \end{array}