Solution:
Let the coefficient matrix be
\[
A = \begin{vmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{vmatrix}, \quad \Delta = \det(A).
\]
Expanding:
\[
\Delta = 2\begin{vmatrix} 2 & 3 \\ -5 & 2 \end{vmatrix} - 4\begin{vmatrix} 1 & 3 \\ 2 & 2 \end{vmatrix} + 2a\begin{vmatrix} 1 & 2 \\ 2 & -5 \end{vmatrix}.
\]
Evaluating each minor:
\[
\begin{vmatrix} 2 & 3 \\ -5 & 2 \end{vmatrix} = 4 + 15 = 19, \quad \begin{vmatrix} 1 & 3 \\ 2 & 2 \end{vmatrix} = 1 \cdot 2 - 3 \cdot 2 = 2 - 6 = -4, \quad \begin{vmatrix} 1 & 2 \\ 2 & -5 \end{vmatrix} = (1)(-5) - (2)(2) = -5 - 4 = -9.
\]
So
\[
\Delta = 2 \cdot 19 - 4 \cdot (-4) + 2a \cdot (-9) = 38 + 16 - 18a = 54 - 18a = 18(3 - a).
\]
For a unique solution, we need \(\Delta \neq 0\), hence \(a \neq 3\).
To determine if the system has infinitely many solutions, we require \(\Delta = 0\) (\(a = 3\)), and also the system must be consistent in its rank conditions. Substituting \(a = 3\) into the equations and analyzing the augmented matrix can lead to constraints on \(b\). One finds that if \(a = 3\) and \(b = 8\), the system has infinitely many solutions (making (1) true).
Checking the other claims shows that (2) and (3) are correct for unique solutions under those specific parameter choices. However, for (4) \(a = 3, b = 6\), it does not provide infinitely many solutions (the system fails to have infinitely many solutions with those values).
Hence, statement (4) is not correct.
If the system of linear equations
2x + y – z = 7
x – 3y + 2z = 1
x + 4y + δz = k, where δ, k ∈ R
has infinitely many solutions, then δ + k is equal to: