We can write the equation as \[\begin{pmatrix} 2 & -2 & 1 \\ 2 & -3 & 2 \\ -1 & 2 & 0 \end{pmatrix} \mathbf{v} = k \mathbf{I} \mathbf{v} = \begin{pmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{pmatrix} \mathbf{v}.\]Then \[\begin{pmatrix} 2 - k & -2 & 1 \\ 2 & -3 - k & 2 \\ -1 & 2 & -k \end{pmatrix} \mathbf{v} = \mathbf{0}.\]This equation has a nonzero vector $ \mathbf{v} $ as a solution if and only if \[\begin{vmatrix} 2 - k & -2 & 1 \\ 2 & -3 - k & 2 \\ -1 & 2 & -k \end{vmatrix} = 0.\]Expanding this determinant, we get \begin{align*} \begin{vmatrix} 2 - k & -2 & 1 \\ 2 & -3 - k & 2 \\ -1 & 2 & -k \end{vmatrix} &= (2 - k) \begin{vmatrix} -3 - k & 2 \\ 2 & -k \end{vmatrix} - (-2) \begin{vmatrix} 2 & 2 \\ -1 & -k \end{vmatrix} + \begin{vmatrix} 2 & -3 - k \\ -1 & 2 \end{vmatrix} \\ &= (2 - k)((-3 - k)(-k) - (2)(2)) -(-2) ((2)(-k) - (2)(-1)) + ((2)(2) - (-3 - k)(-1)) \\ &= -k^3 - k^2 + 5k - 3.\end{align*}Thus, $ k^3 + k^2 - 5k + 3 = 0 $. This equation factors as $ (k - 1)^2 (k + 3) = 0, $ so the possible values of $ k $ are $ \boxed{1, -3} $. Note that for $ k = 1, $ we can take $ \mathbf{v} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, $ and for $ k = -3, $ we can take $ \mathbf{v} = \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix} $.