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 passing through three points \( P_1(0, -5, 4) \), \( P_2(4, -9, 2) \), and \( P_3(4, -8, 4) \) with the coefficient of \( x \) equal to -6, follow these steps:

### Step 1: Vector form of the plane
The general equation of a plane is:
\[
Ax + By + Cz = D
\]
We are given that \( A = -6 \) (the coefficient of \( x \) is -6). So, the equation becomes:
\[
-6x + By + Cz = D
\]

### Step 2: Find two direction vectors
To get the normal vector of the plane, we first find two direction vectors using the three points given. Let's denote these vectors as \( \vec{v_1} \) and \( \vec{v_2} \).

\[
\vec{v_1} = P_2 - P_1 = (4 - 0, -9 - (-5), 2 - 4) = (4, -4, -2)
\]
\[
\vec{v_2} = P_3 - P_1 = (4 - 0, -8 - (-5), 4 - 4) = (4, -3, 0)
\]

### Step 3: Cross product of vectors
The cross product of \( \vec{v_1} \) and \( \vec{v_2} \) gives a normal vector to the plane. We compute the cross product as follows:

\[
\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -4 & -2 \\ 4 & -3 & 0 \end{vmatrix}
\]

Expanding this determinant:

\[
\vec{n} = \mathbf{i} \left( (-4)(0) - (-2)(-3) \right) - \mathbf{j} \left( (4)(0) - (-2)(4) \right) + \mathbf{k} \left( (4)(-3) - (-4)(4) \right)
\]
\[
\vec{n} = \mathbf{i} (0 - 6) - \mathbf{j} (0 + 8) + \mathbf{k} (-12 + 16)
\]
\[
\vec{n} = -6\mathbf{i} - 8\mathbf{j} + 4\mathbf{k}
\]
Thus, the normal vector to the plane is \( \vec{n} = (-6, -8, 4) \).

### Step 4: Equation of the plane
The normal vector \( \vec{n} = (-6, -8, 4) \) gives the coefficients of \( x \), \( y \), and \( z \) in the plane equation. Since we know the coefficient of \( x \) is already \( -6 \), the equation of the plane becomes:
\[
-6x - 8y + 4z = D
\]

### Step 5: Solve for \( D \)
To find \( D \), we use one of the points on the plane, say \( P_1(0, -5, 4) \). Substituting \( x = 0 \), \( y = -5 \), and \( z = 4 \) into the equation:

\[
-6(0) - 8(-5) + 4(4) = D
\]
\[
0 + 40 + 16 = D
\]
\[
D = 56
\]

### Final equation of the plane:
The equation of the plane is:
\[
-6x - 8y + 4z = 56
\]
Or, equivalently:
\[
6x + 8y - 4z = -56
\]

Would you like further clarification or details on any steps? Here are some related questions to deepen your understanding:

1. How do you find the equation of a plane given its normal vector and a point?
2. What are the properties of the cross product between two vectors?
3. Can a plane have more than one normal vector? Why or why not?
4. How does the condition of the coefficient of \( x \) help in simplifying the problem?
5. What is the geometric interpretation of a plane in 3D space?

**Tip**: Always double-check the direction of your normal vector using cross products and ensure that it satisfies the condition for specific coefficients.

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.

This problem involves finding the equation of a plane containing three points, given that the coefficient of x is -6. The solution uses vector algebra, cross products, and the plane equation formula, with a detailed step-by-step explanation. Suitable for Grades 10-12.

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