Let's break down the three parts of the problem based on the image you've uploaded:

Part i)

We are asked to turn the two parametric lines back into linear equations of the form $ax_1 + bx_2 = c$ and solve the system. The given lines are:

$\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + \alpha \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ $\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} + \beta \begin{pmatrix} 2 \\ 1 \end{pmatrix}$

We need to convert these parametric lines into linear equations by eliminating the parameters $\alpha$ and $\beta$.

For the first line: $x_1 = 2 + \alpha \quad \text{and} \quad x_2 = 0 + \alpha = \alpha$ Thus, from the first line, $x_2 = x_1 - 2$.

For the second line: $x_1 = 1 + 2\beta \quad \text{and} \quad x_2 = 3 + \beta$ Thus, from the second line, $x_2 = \frac{x_1 - 1}{2} + 3$.

Now we have the system:

1. $x_2 = x_1 - 2$
2. $x_2 = \frac{x_1 - 1}{2} + 3$

Part ii)

To solve for the interception directly, we equate the two expressions for $x_2$ from the system above:

$x_1 - 2 = \frac{x_1 - 1}{2} + 3$

First, multiply both sides of the equation by 2 to eliminate the fraction: $2(x_1 - 2) = (x_1 - 1) + 6$ $2x_1 - 4 = x_1 + 5$ Now subtract $x_1$ from both sides: $x_1 - 4 = 5$ Solving for $x_1$: $x_1 = 9$

Substitute $x_1 = 9$ into the first equation $x_2 = x_1 - 2$: $x_2 = 9 - 2 = 7$

Thus, the two lines intersect at the point $(9, 7)$.

Part iii)

In the third part, we are given two new parametric lines in 3D space and asked whether they still intersect:

$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix} + \alpha \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ $\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} + \beta \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$

The system becomes: