To find the equation of a plane passing through three points, we use the formula:

$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$

where $(x_1, y_1, z_1)$ is a point on the plane, and $\vec{n} = \langle a, b, c \rangle$ is a normal vector to the plane.

Step 1: Find two vectors on the plane

Given the three points:

$P_1 = (-4, -2, 3), \quad P_2 = (-4, 2, 1), \quad P_3 = (1, 1, -3)$

We can find two vectors lying on the plane by subtracting the coordinates of the points.

1. Vector $\mathbf{v_1} = P_2 - P_1 = (-4 - (-4), 2 - (-2), 1 - 3) = (0, 4, -2)$
2. Vector $\mathbf{v_2} = P_3 - P_1 = (1 - (-4), 1 - (-2), -3 - 3) = (5, 3, -6)$

Step 2: Find the normal vector to the plane

The normal vector $\vec{n}$ is the cross product of $\mathbf{v_1}$ and $\mathbf{v_2}$.

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & -2 \\ 5 & 3 & -6 \end{vmatrix}$$ Calculating the determinant: $$\vec{n} = \hat{i} \left( \begin{vmatrix} 4 & -2 \\ 3 & -6 \end{vmatrix} \right) - \hat{j} \left( \begin{vmatrix} 0 & -2 \\ 5 & -6 \end{vmatrix} \right) + \hat{k} \left( \begin{vmatrix} 0 & 4 \\ 5 & 3 \end{vmatrix} \right)$$ $$= \hat{i} (4(-6) - (-2)(3)) - \hat{j} (0(-6) - (-2)(5)) + \hat{k} (0(3) - 4(5))$$ $$= \hat{i} (-24 + 6) - \hat{j} (0 + 10) + \hat{k} (0 - 20)$$ $$= \hat{i} (-18) - \hat{j} (10) + \hat{k} (-20)$$ Thus, the normal vector $$\vec{n} = (-18, -10, -20)$$. ### Step 3: Write the equation of the plane Using the point $$P_1 = (-4, -2, 3)$$ and the normal vector $$\vec{n} = (-18, -10, -20)$$, we can substitute into the equation of the plane: $$-18(x - (-4)) - 10(y - (-2)) - 20(z - 3) = 0$$ Simplify: $$-18(x + 4) - 10(y + 2) - 20(z - 3) = 0$$ Expand: $$-18x - 72 - 10y - 20 - 20z + 60 = 0$$ $$-18x - 10y - 20z - 32 = 0$$ Thus, the equation of the plane is: $$18x + 10y + 20z = -32$$ ### Final Answer: $$18x + 10y + 20z = -32$$ --- Would you like more details or clarification on any steps? Here are some additional questions for practice: 1. How would the equation of the plane change if the points were different? 2. What is the significance of the normal vector in the context of the plane's equation? 3. How do you check if a point lies on the plane? 4. How can you find the distance from a point to this plane? 5. What would happen if the three points were collinear? **Tip**: Always make sure that the vectors you find are not parallel, as that would mean the points are collinear, and you can’t form a plane!