To determine the equation of a plane, we typically need a point that lies on the plane and a vector that is normal (perpendicular) to the plane. The problem provides specific conditions including two points on the plane and a direction vector of a ray parallel to the plane.

Plane Equation Derivation

The general equation of a plane in Cartesian coordinate system is expressed as:

\[ Ax + By + Cz + D = 0 \]

where \( (A, B, C) \) represent the direction ratios of the normal vector to the plane.

Identifying Given Data

We are provided with the following crucial pieces of information:

• Points on the plane: Let's denote these points as \( P_1 = (0, 6, 0) \) and \( P_2 = (-2, -3, 4) \).
• Parallel Ray Direction: The plane is parallel to a ray whose direction ratios are \( (2, 3, -2) \). We can represent this as a vector \( \vec{v} = \langle 2, 3, -2 \rangle \).

Since both points \( P_1 \) and \( P_2 \) lie on the plane, the vector connecting them, \( \vec{P_1P_2} \), must also lie within the plane. Let's calculate this vector:

\[ \vec{P_1P_2} = P_2 - P_1 = (-2 - 0, -3 - 6, 4 - 0) = \langle -2, -9, 4 \rangle \]

Furthermore, because the given ray is parallel to the plane, its direction vector \( \vec{v} \) is also parallel to the plane. This means that \( \vec{v} \) can be considered as lying within the plane or parallel to a vector within the plane.

Normal Vector Calculation

The normal vector \( \vec{n} = \langle A, B, C \rangle \) to the plane is perpendicular to every vector lying in the plane. Therefore, \( \vec{n} \) must be perpendicular to both \( \vec{P_1P_2} \) and \( \vec{v} \). A straightforward way to find such a normal vector is by computing the cross product of these two vectors:

\[ \vec{n} = \vec{P_1P_2} \times \vec{v} \]

Let's compute the cross product:

\[ \vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & -9 & 4 \\ 2 & 3 & -2 \end{vmatrix} \]

Expanding the determinant:

\[ \mathbf{i}((-9)(-2) - (4)(3)) - \mathbf{j}((-2)(-2) - (4)(2)) + \mathbf{k}((-2)(3) - (-9)(2)) \]

\[ \mathbf{i}(18 - 12) - \mathbf{j}(4 - 8) + \mathbf{k}(-6 + 18) \]

\[ \mathbf{i}(6) - \mathbf{j}(-4) + \mathbf{k}(12) \]

\[ \vec{n} = \langle 6, 4, 12 \rangle \]

The direction ratios of the normal vector are \( (6, 4, 12) \). We can simplify these ratios by dividing by their greatest common divisor, which is 2. This gives us simplified direction ratios \( (3, 2, 6) \). So, we can use \( (A, B, C) = (3, 2, 6) \) for the plane's equation.

Constructing the Plane Equation

Now that we have the coefficients \( A=3, B=2, C=6 \), the partial equation of the plane is:

\[ 3x + 2y + 6z + D = 0 \]

To determine the constant \( D \), we can substitute the coordinates of any given point that lies on the plane into this equation. Let's use point \( P_1 = (0, 6, 0) \):

\[ 3(0) + 2(6) + 6(0) + D = 0 \]

\[ 0 + 12 + 0 + D = 0 \]

\[ 12 + D = 0 \]

\[ D = -12 \]

Finally, substituting the value of \( D \) back into the partial equation, we obtain the complete equation of the plane:

\[ 3x + 2y + 6z - 12 = 0 \]

This equation correctly describes the plane that contains the given points and is parallel to the specified ray.