There exist two distinct unit vectors $\mathbf{v}$ such that the angle between $\mathbf{v}$ and $\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}$ is $45^\circ,$ and the angle between $\mathbf{v}$ and $\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$ is $60^\circ.$  Let $\mathbf{v}_1$ and $\mathbf{v}_2$ be these vectors.  Find $\|\mathbf{v}_1 - \mathbf{v}_2\|.$
Let $\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}.$  Since $\mathbf{v}$ is a unit vector, $x^2 + y^2 + z^2 = 1.$

Since the angle between $\mathbf{v}$ and $\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}$ is $45^\circ,$
\[\frac{2x + 2y - z}{\sqrt{2^2 + 2^2 + (-1)^2}} = \cos 45^\circ = \frac{1}{\sqrt{2}}.\]Then $2x + 2y - z = \frac{3}{\sqrt{2}}.$

Since the angle between $\mathbf{v}$ and $\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$ is $60^\circ,$
\[\frac{y - z}{\sqrt{0^2 + 1^2 + (-1)^2}} = \cos 60^\circ = \frac{1}{2}.\]Then $y - z = \frac{\sqrt{2}}{2}.$

Hence, $y = z + \frac{\sqrt{2}}{2}.$  From the equation $2x + 2y - z = \frac{3}{\sqrt{2}},$
\begin{align*}
x &= -y + \frac{z}{2} + \frac{3}{2 \sqrt{2}} \\
&= -\left( z + \frac{\sqrt{2}}{2} \right) + \frac{z}{2} + \frac{3}{2 \sqrt{2}} \\
&= -\frac{z}{2} + \frac{1}{2 \sqrt{2}}.
\end{align*}Substituting into the equation $x^2 + y^2 + z^2 = 1,$ we get
\[\left( -\frac{z}{2} + \frac{1}{2 \sqrt{2}} \right)^2 + \left( z + \frac{\sqrt{2}}{2} \right)^2 + z^2 = 1.\]This simplifies to $6z^2 + 2z \sqrt{2} - 1 = 0.$  The solutions are $z = \frac{1}{3 \sqrt{2}}$ and $z = -\frac{1}{\sqrt{2}}.$  The possible vectors $\mathbf{v}$ are then
\[\begin{pmatrix} \frac{1}{3 \sqrt{2}} \\ \frac{4}{3 \sqrt{2}} \\ \frac{1}{3 \sqrt{2}} \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ -\frac{1}{\sqrt{2}} \end{pmatrix},\]and the distance between these vectors is $\boxed{\sqrt{2}}.$