There is enough information. Unless two lines are parallel, there is a unique line of minimum length that joins them.
If $\mathbf{U}$ and $\mathbf{V}$ denote the direction vectors of the two given lines, then you are correct that the minimum-length line is in the direction of the cross product $\mathbf{W} = \mathbf{U} \times \mathbf{V}$.
The first line is defined by the point $\mathbf{P} = (1,2,-1)$ and the vector $\mathbf{U}=(1,-1,1)$, so its equation can be written $\mathbf{A}(s) = \mathbf{P} + s\mathbf{U}$. Similarly, the second line has equation $\mathbf{B}(t) = \mathbf{Q}+t\mathbf{V}$, where $\mathbf{Q} = (4,1,0)$ and $\mathbf{V}=(-2,0,-1)$.
To achieve minumum distance, the "joining" line $\mathbf{A}(s) - \mathbf{B}(t)$ must be perpendicular to the given two lines, so:
$$
\left[(\mathbf{A}(s) - \mathbf{B}(t) \right] \cdot \mathbf{U} = 0
$$
$$
\left[(\mathbf{A}(s) - \mathbf{B}(t) \right] \cdot \mathbf{V} = 0
$$
Substituting for $\mathbf{A}(s)$ and $\mathbf{B}(t)$ and rearranging, we get
$$
s(\mathbf{U} \cdot \mathbf{U}) - t(\mathbf{U} \cdot \mathbf{V}) =
\mathbf{U} \cdot (\mathbf{Q} - \mathbf{P})
$$
$$
s(\mathbf{U} \cdot \mathbf{V}) - t(\mathbf{V} \cdot \mathbf{V}) =
\mathbf{V} \cdot (\mathbf{Q} - \mathbf{P})
$$
These equations have a unique solution unless the original two lines are parallel. If $s_0$ and $t_0$ are the solutions, then the closest points of the two lines are then $\mathbf{A}(s_0)$ and $\mathbf{B}(t_0)$.
