Find an equation of a plane containing the three points (0, -5, 4), (4, -9, 2), (4, -8, 4) in which the coefficient of x
 is -6.

To find the equation of a plane that contains three points and has a specific condition (the coefficient of \(x\) is \(-6\)), we can follow these steps:

1. **Determine the normal vector to the plane**:
   A plane in 3D can be expressed as:
   \[
   Ax + By + Cz = D
   \]
   The vector \(\mathbf{n} = (A, B, C)\) is normal (perpendicular) to the plane. To find \(\mathbf{n}\), we compute the cross product of two vectors formed by the given points.

2. **Find two vectors on the plane**:
   Using the three given points \(P_1 = (0, -5, 4)\), \(P_2 = (4, -9, 2)\), and \(P_3 = (4, -8, 4)\), we can form two vectors:
   \[
   \mathbf{v_1} = P_2 - P_1 = (4 - 0, -9 - (-5), 2 - 4) = (4, -4, -2)
   \]
   \[
   \mathbf{v_2} = P_3 - P_1 = (4 - 0, -8 - (-5), 4 - 4) = (4, -3, 0)
   \]

3. **Compute the cross product**:
   The cross product of \(\mathbf{v_1} = (4, -4, -2)\) and \(\mathbf{v_2} = (4, -3, 0)\) will give us a normal vector \(\mathbf{n}\).
   \[
   \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -4 & -2 \\ 4 & -3 & 0 \end{vmatrix}
   \]
   Using the determinant formula:
   \[
   \mathbf{n} = \mathbf{i} \left( (-4)(0) - (-2)(-3) \right) - \mathbf{j} \left( (4)(0) - (-2)(4) \right) + \mathbf{k} \left( (4)(-3) - (4)(-4) \right)
   \]
   Simplifying each component:
   \[
   \mathbf{n} = \mathbf{i} (0 - 6) - \mathbf{j} (0 + 8) + \mathbf{k} (-12 + 16)
   \]
   \[
   \mathbf{n} = (-6, -8, 4)
   \]

4. **Adjust the normal vector**:
   The normal vector we obtained is \(\mathbf{n} = (-6, -8, 4)\). Since we need the coefficient of \(x\) to be \(-6\), this normal vector is already appropriately scaled.

5. **Write the equation of the plane**:
   Using the normal vector \((-6, -8, 4)\) and the general form \(Ax + By + Cz = D\), we substitute the point \(P_1(0, -5, 4)\) to solve for \(D\). Substituting \(x = 0\), \(y = -5\), and \(z = 4\) into the equation:
   \[
   -6(0) - 8(-5) + 4(4) = D
   \]
   \[
   0 + 40 + 16 = D \quad \Rightarrow \quad D = 56
   \]

   Thus, the equation of the plane is:
   \[
   -6x - 8y + 4z = 56
   \]

The equation of the plane is \(-6x - 8y + 4z = 56\).

Would you like to see more details or clarifications?

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Here are 5 related questions to further your understanding:
1. How do you compute the cross product of two vectors in 3D?
2. What is the significance of a plane's normal vector?
3. Can a plane be defined by more than three points?
4. How do you find the distance from a point to a plane?
5. What is the geometric interpretation of the coefficients in a plane equation?

**Tip**: The cross product of two vectors results in a vector perpendicular to both, which is very useful in geometry problems!

Find an equation of a plane containing the three points (0, -5, 4), (4, -9, 2), (4, -8, 4) in which the coefficient of x is -6.

Discover how to find the equation of a plane through the points (0, -5, 4), (4, -9, 2), and (4, -8, 4) with the condition that the coefficient of x is -6. This solution involves vectors, the cross product, and key geometry concepts. Suitable for advanced high school or early college students.

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