The problem consists of two parts regarding the plane $E$ and the line $g$:

Given Plane $E$:

$E: \vec{x} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} + r \cdot \begin{pmatrix} 0 \\ 2 \\ -4 \end{pmatrix} + s \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad r, s \in \mathbb{R}$

1. Equation of the Plane $E$ in Coordinate Form:

We are asked to derive the plane's equation in the form $Ax_1 + Bx_2 + Cx_3 = D$.

We know that the normal vector $\vec{n}$ to the plane is obtained by taking the cross product of the direction vectors: $\vec{v_1} = \begin{pmatrix} 0 \\ 2 \\ -4 \end{pmatrix}, \quad \vec{v_2} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

Perform the cross product $\vec{n} = \vec{v_1} \times \vec{v_2}$: $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -4 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(2 - (-4)) - \hat{j}(0 - (-4)) + \hat{k}(0 - 2)$ $\vec{n} = 6\hat{i} - 4\hat{j} - 2\hat{k} = \begin{pmatrix} 6 \\ -4 \\ -2 \end{pmatrix}$

Thus, the equation of the plane is: $6(x_1 - 1) - 4(x_2 + 2) - 2(x_3 - 3) = 0$ Expanding this: $6x_1 - 6 - 4x_2 - 8 - 2x_3 + 6 = 0$ $6x_1 - 4x_2 - 2x_3 = 8$ Thus, the equation of the plane is: $6x_1 - 4x_2 - 2x_3 = 8$

2. Intersection Point of the Plane $E$ and Line $g$:

The line $g$ is given by: $g: \vec{x} = \begin{pmatrix} -1 \\ -2 \\ -3 \end{pmatrix} + t \cdot \begin{pmatrix} -3 \\ 2 \\ 1 \end{pmatrix}, \quad t \in \mathbb{R}$

To find the intersection point, substitute the parametric form of the line into the equation of the plane. The parametric equations for the line are: $x_1 = -1 - 3t, \quad x_2 = -2 + 2t, \quad x_3 = -3 + t$

Substitute these into the plane equation $6x_1 - 4x_2 - 2x_3 = 8$: $6(-1 - 3t) - 4(-2 + 2t) - 2(-3 + t) = 8$ Expand and simplify: $-6 - 18t + 8 - 8t + 6 - 2t = 8$ $-20t = 8$ $t = -\frac{2}{5}$

Now substitute $t = -\frac{2}{5}$ back into the parametric equations for the line: $x_1 = -1 - 3\left(-\frac{2}{5}\right) = -1 + \frac{6}{5} = -\frac{5}{5} + \frac{6}{5} = \frac{1}{5}$ $x_2 = -2 + 2\left(-\frac{2}{5}\right) = -2 - \frac{4}{5} = -\frac{10}{5} - \frac{4}{5} = -\frac{14}{5}$ $x_3 = -3 + \left(-\frac{2}{5}\right) = -3 - \frac{2}{5} = -\frac{15}{5} - \frac{2}{5} = -\frac{17}{5}$

Thus, the intersection point is: $P\left( \frac{1}{5}, -\frac{14}{5}, -\frac{17}{5} \right)$

Would you like further clarifications or details? Here are some related questions:

1. How do you find the parametric equation of a plane from its point-normal form?
2. How does the cross product relate to the equation of a plane?
3. What is the geometric interpretation of the intersection of a plane and a line?
4. How would the solution change if the line was parallel to the plane?
5. What if the system had no solution, meaning no intersection between the plane and line?

Tip: Always verify the normal vector of the plane by ensuring it is perpendicular to the given direction vectors.