Determine the line $R$ 
Determine the line $R$ that goes throught the point $p$, parallel to the plane $\alpha$ and dissects the line $L$
Given $p=\begin{bmatrix}2 \\ 3 \\ -1 \end{bmatrix}, \alpha \equiv \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}= \begin{bmatrix}0 \\ -1 \\ -1\end{bmatrix} + \rho\begin{bmatrix}1 \\ 2 \\ 2 \end{bmatrix} + \tau\begin{bmatrix}5 \\ 5 \\ 6 \end{bmatrix}$
and $L \equiv \begin{cases} 5x_1 -x_2 -x_3 +2 = 0 \\ 12x_1 -3x_2 -2x_3 +4=0 \end{cases}$

I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $L\equiv \begin{bmatrix}2 \\ 4 \\ 0 \end{bmatrix} +\sigma \begin{bmatrix}1 \\ 1\\ 3 \end{bmatrix}$.
To get the vector for $R$ you have to "scalar multiply" the vector of $L$ and $R$ which need to be equal to $0$. A possibility can be $\begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}$ but then I have to check parallelity with the plane $\alpha$. With the Gauss-Jordan method I checked if the last vector is parallel to the plan $\alpha$ which it is.
So I think $R\equiv\begin{bmatrix}2 \\ 3\\ -1 \end{bmatrix}+ \omicron\begin{bmatrix}0 \\ -3 \\ 1 \end{bmatrix}$

Is this correct?

P.S.: I never had spatial geometry lessons. I had some books so mistakes can be made.

Being parallel to the plane is equivalent to the condition that the line is perpendicular to the plane's normal (vector).

Given $\alpha \equiv \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}= \begin{bmatrix}0 \\ -1 \\ -1\end{bmatrix} + \rho\color{blue}{\begin{bmatrix}1 \\ 2 \\ 2 \end{bmatrix}} + \tau\color{purple}{\begin{bmatrix}5 \\ 5 \\ 6 \end{bmatrix}}$ 

Either converting this to a cartesian equation or computing the following cross product gives you a normal vector of $\alpha$:
$$\vec n_\alpha = \color{blue}{\begin{bmatrix}1 \\ 2 \\ 2 \end{bmatrix}} \times \color{purple}{\begin{bmatrix}5 \\ 5 \\ 6 \end{bmatrix}}=\color{red}{\begin{bmatrix}2 \\ 4 \\ -5 \end{bmatrix}}$$

and $L \equiv \begin{cases} 5x_1 -x_2 -x_3 +2 = 0 \\ 12x_1 -3x_2 -2x_3 +4=0 \end{cases}$
I've determined the vector equation of $L$ by the Gauss-Jordan method which will result in $L\equiv \begin{bmatrix}2 \\ 4 \\ 0 \end{bmatrix} +\sigma \begin{bmatrix}1 \\ 1\\ 3 \end{bmatrix}$. 

Maybe something went wrong in the calculations. A vector equation of $L$ is given by:
$$L\equiv \begin{bmatrix}0 \\ 0 \\ 2 \end{bmatrix} +\sigma \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix} \tag{$*$}$$
Now a direction vector of $R$, through $p$ and a point on $L$, is given by their difference:
$$\underbrace{\begin{bmatrix}0 \\ 0 \\ 2 \end{bmatrix} +\sigma \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}}_{\mbox{point on $L$}}-\underbrace{\begin{bmatrix}2 \\ 3 \\ -1 \end{bmatrix}}_{p}=\color{green}{\begin{bmatrix} \sigma-2 \\ 2\sigma-3 \\ 3\sigma+1 \end{bmatrix}}$$
Recall that we want this to be orthogonal to $\vec n_\alpha$, so:
$$\color{green}{\begin{bmatrix} \sigma-2 \\ 2\sigma-3 \\ 3\sigma+1 \end{bmatrix}} \cdot \color{red}{\begin{bmatrix}2 \\ 4 \\ -5 \end{bmatrix}} = 0 \iff \sigma = \ldots$$
This $\sigma$ gives you the point on $L$ where $R$ passes through and thus, together with $p$, completely determines $R$. So with this value for $\sigma$, you have:
$$R\equiv \begin{bmatrix}2 \\ 3 \\ -1 \end{bmatrix} + \mu \color{green}{\begin{bmatrix} \sigma-2 \\ 2\sigma-3 \\ 3\sigma+1 \end{bmatrix}}$$

Addition after comments. I wouldn't focus on the first coordinate but on the other two:


*

*$R_2 \to R_2-3R_1$

*$R_1 \to R_1+R_2$


$$\begin{bmatrix} 5 & -1 & -1 & -2 \\ 12 & -3 & -2 & -4 \end{bmatrix}
\sim \begin{bmatrix} 5 & -1 & -1 & -2 \\ -3 & 0 & 1 & 2 \end{bmatrix}
\sim \begin{bmatrix} 2 & -1 & 0 & 0 \\ -3 & 0 & 1 & 2 \end{bmatrix}$$
And from here it's easy to see you arrive at $(*)$, solving in terms of the first coordinate.
