For the system of equations to have infinitely many solutions, the determinant of the coefficient matrix (\(\Delta\)) and the determinants of the matrices obtained by replacing each column with the constant terms (\(\Delta_x\), \(\Delta_y\), \(\Delta_z\)) must all be equal to zero. The coefficient matrix is: \[ \begin{pmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{pmatrix} \] \[ \Delta = \begin{vmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{vmatrix} = 1(15 - 4) - 1(6 - 2) + a(4 - 5) = 11 - 4 - a = 7 - a. \] For infinitely many solutions, \(\Delta = 0\), so \(7 - a = 0 \Rightarrow a = 7\). Now, let's calculate \(\Delta_x\): \[ \Delta_x = \begin{vmatrix} b & 1 & 7 \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{vmatrix} = b(15 - 4) - 1(18 - 6) + 7(12 - 15) = 11b - 12 - 21 = 11b - 33. \] For infinitely many solutions, \(\Delta_x = 0\), so \(11b - 33 = 0 \Rightarrow b = 3\). Now we can calculate \(2a + 3b = 2(7) + 3(3) = 14 + 9 = 23\).
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: