Is there an isometry $\mathbb R^3\to \mathbb R^3 $ with this property? What is an isometry $\mathbb R^3\to \mathbb R^3 $ which maps $(3,1,2) \mapsto (2,2,2)$ and $(2,2,2) \mapsto (1,1,2)$?
Hint: the first map can be handled by a simple translation.  Once you've factored out that translation, consider the 'relative positions' of the points in the second map (i.e., relative to their corresponding points in the first map) - these would be $(2,2,2)-(3,1,2) = (-1, 1, 0)$ and $(1,1,2) - (2,2,2) = (-1, -1, 0)$; can you find a rotation that maps one to the other, or prove that there isn't one?  And once you've got that, can you see how all the transformations can be composed to find a final result?
Isometries should preserve the length, but $(3,1,2)$ and $(2,2,2)$ have different length (as well as $(2,2,2)$ and $(1,1,2)$. Hence, there is no such isometry.