Step 1: Parametric equations of the first line:
The parametric equations of the first line are derived from: \[ \frac{x-k}{2} = \frac{y+1}{3} = \frac{z-1}{4} = t. \] This yields: \[ x = 2t + k, \quad y = 3t - 1, \quad z = 4t + 1. \]
Step 2: Parametric equations of the second line:
For the second line, the parametric equations are obtained from: \[ \frac{x-3}{1} = \frac{y - 9/2}{2} = z = s. \]
Step 3: Condition for intersection:
For the two lines to intersect, their parametric coordinates must be equal: \[ x_1 = x_2, \quad y_1 = y_2, \quad z_1 = z_2. \] Substituting the equations gives: \[ 2t + k = s + 3, \quad 3t - 1 = 2s + \frac{9}{2}, \quad 4t + 1 = s. \]
Step 4: Solve the system of equations:
From the third equation, \( 4t + 1 = s \), we substitute this into the first equation: \[ 2t + k = (4t + 1) + 3. \] Simplifying leads to: \[ 2t + k = 4t + 4. \] Solving for \( k \): \[ k = 4t + 4 - 2t = 4 - 2t. \] Now, substitute \( s = 4t + 1 \) into the second equation: \[ 3t - 1 = 2(4t + 1) + \frac{9}{2}. \] Simplifying: \[ 3t - 1 = 8t + 2 + \frac{9}{2}. \] Combining terms: \[ 3t - 1 = 8t + \frac{13}{2}. \] Multiplying by 2 to clear the fraction: \[ 6t - 2 = 16t + 13. \] Rearranging to solve for \( t \): \[ 16t - 6t = -2 - 13. \] \[ 10t = -15 \implies t = -\frac{3}{2}. \]
Step 5: Find \( k \):
Substitute \( t = -\frac{3}{2} \) into the expression for \( k \): \[ k = 4 - 2\left(-\frac{3}{2}\right). \] Simplifying: \[ k = 4 + 3 = 7. \]
Final Answer: \[ \boxed{7}. \]