To determine if the lines are skew, we need to check if they are neither parallel nor intersecting.

First, equate the components of the lines:

For the i component: \(2 + 2s = 1 + t\)

For the j component: \(-1 + 3s = 3 + 2t\)

For the k component: \(1 - s = 4 + t\)

Solving these equations:

From \(2 + 2s = 1 + t\), we get \(t = 2s - 1\).

From \(-1 + 3s = 3 + 2t\), substitute \(t = 2s - 1\):

\(-1 + 3s = 3 + 2(2s - 1)\)

\(-1 + 3s = 3 + 4s - 2\)

\(-1 + 3s = 1 + 4s\)

\(-2 = s\)

Substitute \(s = -2\) into \(t = 2s - 1\):

\(t = 2(-2) - 1 = -4 - 1 = -5\)

Check the k component: \(1 - s = 4 + t\)

\(1 - (-2) = 4 + (-5)\)

\(3 \neq -1\)

The equations do not satisfy all components simultaneously, so the lines do not intersect.

Since the direction vectors \((2, 3, -1)\) and \((1, 2, 1)\) are not scalar multiples, the lines are not parallel.

Therefore, the lines are skew.