Step-by-step Solution:

To determine the correct statement(s) about the system of linear equations: \[ \begin{cases} 2x + 2y + z = 1 \\ 4x + ky + 2z = 2 \\ kx + 4y + z = 1 \end{cases} \] we analyze the system using the determinant of the coefficient matrix. The coefficient matrix \( A \) is: \[ A = \begin{bmatrix} 2 & 2 & 1 \\ 4 & k & 2 \\ k & 4 & 1 \end{bmatrix} \] The determinant of \( A \) is: \[ \det(A) = 2(k \cdot 1 - 2 \cdot 4) - 2(4 \cdot 1 - 2 \cdot k) + 1(4 \cdot 4 - k \cdot k) \] \[ \det(A) = 2(k - 8) - 2(4 - 2k) + (16 - k^2) \] \[ \det(A) = 2k - 16 - 8 + 4k + 16 - k^2 \] \[ \det(A) = -k^2 + 6k - 8 \] \[ \det(A) = -(k^2 - 6k + 8) = -(k - 2)(k - 4) \] The determinant is zero when \( k = 2 \) or \( k = 4 \). \[1. Unique Solution:\] - If \( \det(A) \neq 0 \), the system has a unique solution. - This occurs when \( k \neq 2 \) and \( k \neq 4 \). \[2. Infinite Solutions:\] - If \( \det(A) = 0 \) and the system is consistent, it has infinite solutions. - For \( k = 4 \), substituting into the equations shows consistency, leading to infinite solutions. - For \( k = 2 \), substituting into the equations shows inconsistency, so no infinite solutions. \[3. Inconsistent System:\] - If \( \det(A) = 0 \) and the system is inconsistent, it has no solution. - For \( k = 2 \), the system is inconsistent. Correct Statements: - (A) The system has a unique solution if \( k \neq 4 \) and \( k \neq 2 \). - (C) The system has infinite solutions if \( k = 4 \). - (D) The system has infinite solutions if \( k = 2 \) is incorrect; it is inconsistent. Final Answer: \[ \boxed{C} \]