1. **State the problem:** Find the vector and Cartesian equations of the plane passing through points $A(-2,-2,2)$, $B(3,2,3)$, and $C(2,-2,2)$. 2. **Formula and rules:** - The vector equation of a plane through point $\mathbf{r_0}$ with normal vector $\mathbf{n}$ is $\mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0$. - The Cartesian equation is derived from the vector form and is $Ax + By + Cz + D = 0$ where $\mathbf{n} = (A,B,C)$. - To find $\mathbf{n}$, compute the cross product of two vectors lying on the plane. 3. **Find vectors on the plane:** $$\vec{AB} = B - A = (3 - (-2), 2 - (-2), 3 - 2) = (5, 4, 1)$$ $$\vec{AC} = C - A = (2 - (-2), -2 - (-2), 2 - 2) = (4, 0, 0)$$ 4. **Find the normal vector $\mathbf{n}$:** $$\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 4 & 1 \\ 4 & 0 & 0 \end{vmatrix} = (4 \cdot 0 - 1 \cdot 0)\mathbf{i} - (5 \cdot 0 - 1 \cdot 4)\mathbf{j} + (5 \cdot 0 - 4 \cdot 4)\mathbf{k} = (0)\mathbf{i} - (-4)\mathbf{j} - (16)\mathbf{k} = (0,4,-16)$$ 5. **Simplify the normal vector:** Divide by 4: $$\mathbf{n} = (0,1,-4)$$ 6. **Vector equation of the plane:** Using point $A(-2,-2,2)$: $$\mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0 \Rightarrow (0,1,-4) \cdot ((x,y,z) - (-2,-2,2)) = 0$$ $$0(x+2) + 1(y+2) - 4(z-2) = 0$$ 7. **Cartesian equation:** Simplify: $$y + 2 - 4z + 8 = 0 \Rightarrow y - 4z + 10 = 0$$ **Final answers:** - Vector equation: $$\mathbf{r} = (-2,-2,2) + s(5,4,1) + t(4,0,0)$$ where $s,t \in \mathbb{R}$. - Cartesian equation: $$y - 4z + 10 = 0$$