1. **Problem statement:** Given points $A(3,0,0)$, $B(1,2,-3)$, and $C(0,1,-1)$, find the coordinates of point $D$ such that $ABCD$ is a parallelogram, and then find the area of parallelogram $ABCD$. 2. **Formula for coordinates of $D$:** In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal $AC$ equals the midpoint of diagonal $BD$. Using this, the coordinates of $D$ can be found by: $$D = A + C - B$$ 3. **Calculate coordinates of $D$:** $$D_x = 3 + 0 - 1 = 2$$ $$D_y = 0 + 1 - 2 = -1$$ $$D_z = 0 + (-1) - (-3) = 0 + (-1) + 3 = 2$$ So, $D = (2, -1, 2)$. 4. **Formula for area of parallelogram:** The area of parallelogram $ABCD$ is the magnitude of the cross product of vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$: $$\text{Area} = \| \overrightarrow{AB} \times \overrightarrow{AD} \|$$ 5. **Calculate vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$:** $$\overrightarrow{AB} = B - A = (1 - 3, 2 - 0, -3 - 0) = (-2, 2, -3)$$ $$\overrightarrow{AD} = D - A = (2 - 3, -1 - 0, 2 - 0) = (-1, -1, 2)$$ 6. **Calculate cross product $\overrightarrow{AB} \times \overrightarrow{AD}$:** $$\overrightarrow{AB} \times \overrightarrow{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 2 & -3 \\ -1 & -1 & 2 \end{vmatrix} = \mathbf{i}(2 \cdot 2 - (-3) \cdot (-1)) - \mathbf{j}(-2 \cdot 2 - (-3) \cdot (-1)) + \mathbf{k}(-2 \cdot (-1) - 2 \cdot (-1))$$ $$= \mathbf{i}(4 - 3) - \mathbf{j}(-4 - 3) + \mathbf{k}(2 + 2) = \mathbf{i}(1) - \mathbf{j}(-7) + \mathbf{k}(4) = (1, 7, 4)$$ 7. **Calculate magnitude of cross product:** $$\| (1,7,4) \| = \sqrt{1^2 + 7^2 + 4^2} = \sqrt{1 + 49 + 16} = \sqrt{66}$$ 8. **Final answer:** - Coordinates of $D$ are $\boxed{(2, -1, 2)}$. - Area of parallelogram $ABCD$ is $\boxed{\sqrt{66}}$.