Let $ \mathbf{a} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}, $ $ \mathbf{b} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix}, $ $ \mathbf{c} = \begin{pmatrix} 4 \\ 3 \\ 6 \end{pmatrix}, $ and $ \mathbf{d} = \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix} $. First, we find the plane containing $ B, $ $ C, $ and $ D $. The normal vector to this plane is \[(\mathbf{c} - \mathbf{b}) \times (\mathbf{d} - \mathbf{b}) = \begin{pmatrix} 1 \\ 3 \\ 5 \end{pmatrix} \times \begin{pmatrix} -1 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} -12 \\ -6 \\ 6 \end{pmatrix}.\]Scaling, we can take $ \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} $ as the normal vector. Then the equation of the plane is of the form \[2x + y - z + k = 0.\]Substituting any of the coordinates of $ B, $ $ C, $ or $ D, $ we get that the equation of the plane is \[2x + y - z - 5 = 0.\]Then the distance from $ A $ to plane $ BCD $ (acting as the height of the tetrahedron) is \[\frac{|(2)(0) + (1)(1) - (1)(2) - 5|}{\sqrt{2^2 + 1^2 + (-1)^2}} = \frac{6}{\sqrt{6}} = \sqrt{6}.\]The area of triangle $ BCD $ (acting as the base of the tetrahedron) is given by \[\frac{1}{2} \| (\mathbf{c} - \mathbf{b}) \times (\mathbf{d} - \mathbf{b}) \| = \frac{1}{2} \left\| \begin{pmatrix} -12 \\ -6 \\ 6 \end{pmatrix} \right\| = 3 \sqrt{6}.\]Therefore, the volume of tetrahedron $ ABCD $ is \[\frac{1}{3} \cdot 3 \sqrt{6} \cdot \sqrt{6} = \boxed{6}.\]