1. **Statement:** Determine if the lines \(r\) and \(s\) are concurrent (intersect at a point). 2. **Given:** Line \(r\): \(x = 2 - 4t, y = -t, z = -1 + 2t\) Line \(s\): \(x = 2\lambda, y = 1 + \lambda, z = -1 - \lambda\) where \(t, \lambda \in \mathbb{R}\). 3. **To check concurrency:** We need to find if there exist \(t\) and \(\lambda\) such that \[ 2 - 4t = 2\lambda, \quad -t = 1 + \lambda, \quad -1 + 2t = -1 - \lambda \] 4. **From the second equation:** \[ -t = 1 + \lambda \implies \lambda = -t - 1 \] 5. **Substitute \(\lambda = -t - 1\) into the first equation:** \[ 2 - 4t = 2(-t - 1) = -2t - 2 \] Simplify: \[ 2 - 4t = -2t - 2 \] Add \(2t\) to both sides: \[ 2 - 4t + 2t = -2 \implies 2 - 2t = -2 \] Subtract 2 from both sides: \[ -2t = -4 \] Divide both sides by \(-2\): \[ \cancel{-2}t = \frac{-4}{\cancel{-2}} \implies t = 2 \] 6. **Find \(\lambda\) using \(t=2\):** \[ \lambda = -2 - 1 = -3 \] 7. **Check the third equation with \(t=2\) and \(\lambda=-3\):** \[ -1 + 2(2) = -1 - (-3) \implies -1 + 4 = -1 + 3 \implies 3 = 2 \] This is false, so the lines do not intersect. 8. **Conclusion:** The lines \(r\) and \(s\) are not concurrent (they do not intersect).