1. **Problem:** Find the equation of the plane through the points (3, 0, -1), (-2, -2, 3), and (7, 1, -4). 2. **Formula and rules:** The equation of a plane through three points $P_1$, $P_2$, and $P_3$ can be found using the vector form: $$\vec{n} \cdot (\vec{r} - \vec{r_0}) = 0$$ where $\vec{n}$ is the normal vector to the plane, $\vec{r} = (x,y,z)$ is a general point on the plane, and $\vec{r_0}$ is a position vector to one of the points. 3. **Find vectors on the plane:** $$\vec{v_1} = P_2 - P_1 = (-2 - 3, -2 - 0, 3 - (-1)) = (-5, -2, 4)$$ $$\vec{v_2} = P_3 - P_1 = (7 - 3, 1 - 0, -4 - (-1)) = (4, 1, -3)$$ 4. **Find the normal vector $\vec{n}$ by cross product:** $$\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & -2 & 4 \\ 4 & 1 & -3 \end{vmatrix}$$ Calculate determinant: $$\mathbf{i}((-2)(-3) - 4(1)) - \mathbf{j}((-5)(-3) - 4(4)) + \mathbf{k}((-5)(1) - (-2)(4))$$ $$= \mathbf{i}(6 - 4) - \mathbf{j}(15 - 16) + \mathbf{k}(-5 + 8) = 2\mathbf{i} + 1\mathbf{j} + 3\mathbf{k}$$ So, $\vec{n} = (2, 1, 3)$. 5. **Write the plane equation:** Using point $P_1 = (3,0,-1)$: $$2(x - 3) + 1(y - 0) + 3(z + 1) = 0$$ Simplify: $$2x - 6 + y + 3z + 3 = 0$$ $$2x + y + 3z - 3 = 0$$ **Final answer:** $$\boxed{2x + y + 3z = 3}$$