1. **State the problem:** We are given points A(-1,1,2), B(-1,3,0), C(1,1,3), and D(3,3,5) in 3D space. We need to determine if (a) the volume of tetrahedron ABCD is 2, (b) the points are coplanar, or (c) none of the above. 2. **Calculate vectors:** $$\vec{AB} = B - A = (-1 - (-1), 3 - 1, 0 - 2) = (0, 2, -2)$$ $$\vec{AC} = C - A = (1 - (-1), 1 - 1, 3 - 2) = (2, 0, 1)$$ $$\vec{AD} = D - A = (3 - (-1), 3 - 1, 5 - 2) = (4, 2, 3)$$ 3. **Compute the scalar triple product:** The volume $V$ of tetrahedron ABCD is given by $$V = \frac{1}{6} |\vec{AB} \cdot (\vec{AC} \times \vec{AD})|$$ First, compute the cross product $\vec{AC} \times \vec{AD}$: $$\vec{AC} \times \vec{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & 1 \\ 4 & 2 & 3 \end{vmatrix} = \mathbf{i}(0 \cdot 3 - 1 \cdot 2) - \mathbf{j}(2 \cdot 3 - 1 \cdot 4) + \mathbf{k}(2 \cdot 2 - 0 \cdot 4)$$ $$= \mathbf{i}(0 - 2) - \mathbf{j}(6 - 4) + \mathbf{k}(4 - 0) = (-2, -2, 4)$$ 4. **Dot product $\vec{AB} \cdot (\vec{AC} \times \vec{AD})$:** $$\vec{AB} \cdot (-2, -2, 4) = 0 \cdot (-2) + 2 \cdot (-2) + (-2) \cdot 4 = 0 - 4 - 8 = -12$$ 5. **Calculate volume:** $$V = \frac{1}{6} | -12 | = \frac{12}{6} = 2$$ 6. **Interpretation:** Since the volume is 2 (non-zero), the points are not coplanar. **Final answers:** - (a) The volume of tetrahedron ABCD is 2: **True** - (b) The points are coplanar: **False** - (c) None of the above: **False**