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To find the equation of the plane passing through the points \((-2, 6, -8)\), \((-6, 1, 7)\), and \((-2, -5, -8)\), we can follow these steps:

### Step 1: Find two vectors that lie on the plane
The first vector \(\mathbf{v}_1\) is formed by subtracting the coordinates of the first point from the second point:
\[
\mathbf{v}_1 = (-6, 1, 7) - (-2, 6, -8) = (-6 + 2, 1 - 6, 7 + 8) = (-4, -5, 15)
\]

The second vector \(\mathbf{v}_2\) is formed by subtracting the coordinates of the first point from the third point:
\[
\mathbf{v}_2 = (-2, -5, -8) - (-2, 6, -8) = (0, -11, 0)
\]

### Step 2: Find the cross product of these two vectors
The normal vector \(\mathbf{n}\) to the plane is the cross product of \(\mathbf{v}_1\) and \(\mathbf{v}_2\):
\[
\mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & -5 & 15 \\ 0 & -11 & 0 \end{vmatrix}
\]
This gives:
\[
\mathbf{n} = \mathbf{i} \left((-5)(0) - (15)(-11)\right) - \mathbf{j} \left((-4)(0) - (15)(0)\right) + \mathbf{k} \left((-4)(-11) - (-5)(0)\right)
\]
\[
\mathbf{n} = \mathbf{i}(0 + 165) - \mathbf{j}(0) + \mathbf{k}(44)
\]
\[
\mathbf{n} = 165\mathbf{i} + 44\mathbf{k}
\]
Thus, the normal vector is \(\mathbf{n} = (165, 0, 44)\).

### Step 3: Equation of the plane
The equation of the plane is given by:
\[
A(x - x_1) + B(y - y_1) + C(z - z_1) = 0
\]
where \((A, B, C)\) is the normal vector and \((x_1, y_1, z_1)\) is any point on the plane. Using the normal vector \((165, 0, 44)\) and the point \((-2, 6, -8)\), the equation becomes:
\[
165(x + 2) + 0(y - 6) + 44(z + 8) = 0
\]
Simplifying:
\[
165(x + 2) + 44(z + 8) = 0
\]
\[
165x + 330 + 44z + 352 = 0
\]
\[
165x + 44z + 682 = 0
\]

Thus, the equation of the plane is:
\[
165x + 44z + 682 = 0
\]

Let me know if you'd like more details or have any questions! 

Here are some follow-up questions:
1. How would you find the distance from a point to this plane?
2. Can you find the parametric form of the line that lies on this plane?
3. How would the equation change if one of the points were altered slightly?
4. Can you describe the geometric significance of the normal vector?
5. What is the interpretation of the cross product in the context of vectors?

**Tip:** The normal vector to a plane is perpendicular to every vector that lies on the plane.

Find a plane through the points (-2, 6, -8), (-6, 1, 7), (-2, -5, -8).

Learn how to find the equation of a plane passing through the points (-2, 6, -8), (-6, 1, 7), and (-2, -5, -8). This problem involves key concepts such as vector subtraction, cross product, and the plane equation. Suitable for students in Grades 10-12.

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