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.