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The given problem provides a plane in parametric form, and the goal is to find its equation in general form. The parametric form of the plane is:

\[
\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix} + t \begin{pmatrix} 2 \\ -3 \\ 0 \end{pmatrix} + s \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
\]

### Step 1: Finding Two Direction Vectors
The parametric equation describes a plane defined by a point and two direction vectors. We have:

- Point \( P_0 = (4, 1, -2) \)
- Direction vector \( \mathbf{d}_1 = \begin{pmatrix} 2 \\ -3 \\ 0 \end{pmatrix} \)
- Direction vector \( \mathbf{d}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \)

### Step 2: Calculating the Normal Vector
The normal vector \( \mathbf{n} \) to the plane is found by taking the cross product of the two direction vectors \( \mathbf{d}_1 \) and \( \mathbf{d}_2 \):

\[
\mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 0 \\ 1 & 1 & 1 \end{vmatrix}
\]

Calculating the determinant:

\[
\mathbf{n} = \mathbf{i} \begin{vmatrix} -3 & 0 \\ 1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix}
\]

\[
\mathbf{n} = \mathbf{i}(-3 - 0) - \mathbf{j}(2 - 0)

Givet är planet i parameterform: (x, y, z) = (4, 1, -2) + t(2, -3, 0) + s(1, 1, 1). Ange planets ekvation i allmän form (Ax + By + Cz + D = 0).

This problem involves converting the parametric equation of a plane into its general form. It uses vector geometry and the cross product of two direction vectors to find the normal vector. Suitable for students in Grades 11-12 or undergraduate level.

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