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?

---

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!