Step-by-step Solution:

To possess a non-zero solution, we need to find when the determinant of the coefficient matrix is zero: \[ \left| \begin{array}{ccc} 4 & k & 2 \\ k & 4 & 1 \\ 2 & 2 & 1 \end{array} \right| = 0 \] Expanding the determinant: \[ 4 \cdot \left| \begin{array}{cc} 4 & 1 \\ 2 & 1 \end{array} \right| - k \cdot \left| \begin{array}{cc} k & 1 \\ 2 & 1 \end{array} \right| + 2 \cdot \left| \begin{array}{cc} k & 4 \\ 2 & 2 \end{array} \right| \] \[ = 4 \cdot (4 \cdot 1 - 1 \cdot 2) - k \cdot (k \cdot 1 - 1 \cdot 2) + 2 \cdot (k \cdot 2 - 4 \cdot 2) \] \[ = 4 \cdot (4 - 2) - k \cdot (k - 2) + 2 \cdot (2k - 8) \] \[ = 4 \cdot 2 - k(k - 2) + 2 \cdot (2k - 8) \] \[ = 8 - k^2 + 2k + 4k - 16 \] \[ = -k^2 + 6k - 8 \] Now, for a non-zero solution, the determinant must be zero: \[ -k^2 + 6k - 8 = 0 \] Multiplying through by -1: \[ k^2 - 6k + 8 = 0 \] Now, solving the quadratic equation using the quadratic formula: \[ k = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)} \] \[ k = \frac{6 \pm \sqrt{36 - 32}}{2} \] \[ k = \frac{6 \pm \sqrt{4}}{2} \] \[ k = \frac{6 \pm 2}{2} \] Thus, the two possible values of \( k \) are: \[ k = \frac{6 + 2}{2} = 4 \quad \text{and} \quad k = \frac{6 - 2}{2} = 2 \] Therefore, there are two possible values of \( k \), namely \( k = 2 \) and \( k = 4 \).