Express the general points on lines l and m in component form:

Line l: \((1 + 2s, s, -2 + 3s)\)

Line m: \((6 + t, -5 - 2t, 4 + t)\)

Equate the components to find \(s\) and \(t\):

1. \(1 + 2s = 6 + t\)

2. \(s = -5 - 2t\)

3. \(-2 + 3s = 4 + t\)

From equation (1):

\(t = 2s - 5\)

Substitute \(t = 2s - 5\) into equation (2):

\(s = -5 - 2(2s - 5)\)

\(s = -5 - 4s + 10\)

\(5s = 5\)

\(s = 1\)

Substitute \(s = 1\) into \(t = 2s - 5\):

\(t = 2(1) - 5 = -3\)

Verify with equation (3):

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

\(1 = 1\)

All equations are satisfied. The position vector of the intersection point is:

\(3\mathbf{i} + \mathbf{j} + \mathbf{k}\)