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)