1. **Problem Statement:** Determine if the statements about the two lines \(r\) and \(s\) in 3D space are true or false. Given: \[ r: \begin{cases} x=2-4t \\ y=-t \\ z=-1+2t \end{cases}, \quad s: \begin{cases} x=2\lambda \\ y=1+\lambda \\ z=-1-\lambda \end{cases}, \quad t, \lambda \in \mathbb{R} \] 2. **Step 1: Write parametric equations and direction vectors.** - Line \(r\) point: \(P_r=(2,0,-1)\), direction vector \(\vec{v_r}=(-4,-1,2)\). - Line \(s\) point: \(P_s=(0,1,-1)\), direction vector \(\vec{v_s}=(2,1,-1)\). 3. **Step 2: Check if lines intersect.** Set \(r(t) = s(\lambda)\): \[ \begin{cases} 2 - 4t = 2\lambda \\ -t = 1 + \lambda \\ -1 + 2t = -1 - \lambda \end{cases} \] From the third equation: \[ -1 + 2t = -1 - \lambda \implies 2t = -\lambda \implies \lambda = -2t \] Substitute \(\lambda = -2t\) into the first two equations: \[ 2 - 4t = 2(-2t) = -4t \implies 2 - 4t = -4t \] Simplify: \[ 2 = 0 \quad \text{(contradiction)} \] Since this is false, the lines do not intersect. 4. **Step 3: Check if lines are parallel.** Check if \(\vec{v_r}\) is a scalar multiple of \(\vec{v_s}\): \[ \vec{v_r} = (-4,-1,2), \quad \vec{v_s} = (2,1,-1) \] Try to find \(k\) such that \(\vec{v_r} = k \vec{v_s}\): \[ -4 = 2k \implies k = -2 \] Check other components: \[ -1 \stackrel{?}{=} 1 \times (-2) = -2 \quad \text{(false)} \] So, vectors are not parallel. 5. **Step 4: Check if lines are skew (neither parallel nor intersecting).** Since lines do not intersect and are not parallel, they are skew lines. **Final conclusion:** - Lines \(r\) and \(s\) do not intersect. - Lines \(r\) and \(s\) are not parallel. - Therefore, lines \(r\) and \(s\) are skew lines. **Answer:** \[ \text{Lines } r \text{ and } s \text{ are skew lines.} \]