The given complex is
\[[\text{Fe}(\text{NH}_3)_2(\text{CN})_4]^{3+}\]
Here, the oxidation state of iron is \(+3\), which corresponds to a \(d^5\) electronic configuration in the high-spin state. The ligands cyanide (\(\text{CN}^-\)) and ammonia (\(\text{NH}_3\)) are arranged such that the \(e_g\) orbitals remain unoccupied.
Given values:
\(x = 2 \, (number\ of \text{NH}_3 \, \text{ligands})\),
\(y = 4 \, (\text{number of } \text{CN}^- \, \text{ligands})\)
Thus, \(x + y = 2 + 4 = 6\).
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: