1. **Problem statement:** Given three points $A$, $B$, and $C$ in 3D space, find the equation of the plane $E$ passing through these points in coordinate form. 2. **Formula and rules:** The plane equation can be found using the point-normal form: $$\vec{n} \cdot (\vec{x} - \vec{A}) = 0$$ where $\vec{n}$ is the normal vector to the plane, and $\vec{A}$ is the position vector of point $A$. 3. To find $\vec{n}$, compute the cross product of vectors $\vec{AB}$ and $\vec{AC}$: $$\vec{n} = \vec{AB} \times \vec{AC}$$ 4. **Step-by-step for part (a):** - Points: $A(1,1,1)$, $B(1,0,1)$, $C(0,1,1)$ - Compute vectors: $$\vec{AB} = (1-1, 0-1, 1-1) = (0, -1, 0)$$ $$\vec{AC} = (0-1, 1-1, 1-1) = (-1, 0, 0)$$ - Compute cross product: $$\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{vmatrix} = (0 \cdot 0 - 0 \cdot (-1))\mathbf{i} - (0 \cdot 0 - (-1) \cdot (-1))\mathbf{j} + (0 \cdot 0 - (-1) \cdot (-1))\mathbf{k} = (0)\mathbf{i} - (0 - 1)\mathbf{j} + (0 - 1)\mathbf{k} = (0, -1, -1)$$ - Simplify normal vector: $$\vec{n} = (0, -1, -1)$$ - Plane equation: $$0(x - 1) -1(y - 1) -1(z - 1) = 0$$ - Expand: $$-y + 1 - z + 1 = 0$$ $$-y - z + 2 = 0$$ - Multiply both sides by $-1$: $$\cancel{-}y - \cancel{-}z + \cancel{2} = \cancel{0} \Rightarrow y + z = 2$$ **Final equation for (a):** $$y + z = 2$$ 5. Since the user asked to solve the task, we provide the solution for part (a) only as per instructions.