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.
 A: 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.
A: Here are a couple of alternative approaches. Note that the pair of equations that define $L$ are the equations of planes, that is, $L$ is described as the intersection of two planes.  
For $R$ to be parallel to $\alpha$, its direction vector must be some nontrivial linear combination of the generators of $\alpha$, so we know that a parametric equation of $R$ is $p+\lambda(\rho[1,2,2]^T+\tau[5,5,6]^T)$ for some as yet undetermined $\rho$ and $\tau$. The other condition on $R$ is that it intersects $L$, i.e., that it has a common intersection with both of the defining planes. Substituting into their equations produces the system $$\lambda(\rho+14\tau)+10 = 0 \\ \lambda(2\rho+33\tau)+21 = 0.$$ This system is underdetermined, which makes sense because any nonzero scalar multiple of a line’s direction vector is also a direction vector for the line. Set $\lambda$ to some convenient value such as $1$ and solve the resulting system for $\rho$ and $\tau$.  
What you’re essentially doing above is finding the intersection of three planes: the two that define $L$ and a plane through $p$ that is parallel to $\alpha$. The latter plane consists of all of the lines through $p$ that are parallel to $\alpha$. You could use this idea to solve the problem is a slightly different way, which I’ll illustrate using homogeneous coordinates. If the equation of a plane is $ax+by+cz+d=0$, it can be represented by the homogeneous vector $\mathbf\pi = [a,b,c,d]^T$, so that the equation of the plane becomes $\mathbf\pi^T\mathbf x = 0$ (i.e., the dot product of the plane vector and coordinates of a point on the plane). The parallel plane to $\alpha$ through $p$ is then a null vector $\mathbf\pi$ of the matrix $$\begin{bmatrix}2&3&-1&1\\1&2&2&0\\5&5&6&0\end{bmatrix}.$$ (Think about how this relates to the condition $\mathbf\pi^T\mathbf x=0$.) The rows of this matrix are just the point $p$ and the two generators of $\alpha$—in projective-geometric terms, these are three points on the plane, two of which are at infinity. You can compute this null vector in any of the usual ways, such as Gaussian elimination. If you’ve done it correctly, the resulting space should be one-dimensional.† One possibility is $\mathbf\pi = [2,4,-5,-21]^T$. To compute the intersection of $\pi$ and $L$, use the same procedure: assemble the vectors that represent the three planes into a matrix and find a null vector of it, then convert to inhomogeneous Cartesian coordinates by dividing through by the last component of the vector. This gives you another point on $R$, from which I trust you know how to generate an equation of this line.  
† You can also compute the plane’s representative vector by taking the cross product of the two generators of $\alpha$ to get the first three components of $\mathbf\pi$ and setting its last component to the negative of the dot product of this normal vector with $p$. This is a direct analogue of one of the ways you’d find the point-normal equation of the plane. However, this method is specific to three-dimensional space, while the null space method works in spaces of any dimension.
