I must to determinate equation of the plan that containing the line $r= \begin {cases} x=z+1 \\ y=2z+1\end{cases}$
That passes from O$(0,0,0)$ and parallel to the line $r_1= \begin{cases} x=2z+3\\ y=z+5\end{cases}$
Now I'm going to solve the first. $(ax+by+cz+d)+\lambda(ax+by+cz+d)=0, x-1+\lambda y-\lambda - (2\lambda +1)z=0$.
Now I'm doing the passage through the origin $(0,0,0)$ and I'm doing the substitution of $x, y, z$ to $0,0,0$. So I have $\lambda =-1$. For finding the parallel to the r1 I'll going to research the parameters directors and from the matrix $\begin{bmatrix} 1 & 0 & - 2 \\ 0 & 1 & - 1 \end{bmatrix}$ I've got $l=1, m=1, n=1$. For the formula of the line parallel to the plan I've got $al+bm+cn=0$ and here I'm in stuck. I don't know were I must to do the substitution.
We have with $p = (x,y,z)^{\dagger}$
$$ r \to p = \left(\begin{array}{c}1\\1\\0\end{array}\right)+\lambda \left(\begin{array}{c}1\\2\\1\end{array}\right)=p_0+\lambda \vec v_0\\ r_1 \to p = \left(\begin{array}{c}3\\5\\0\end{array}\right)+\lambda_1 \left(\begin{array}{c}2\\1\\1\end{array}\right)=p_1+\lambda_1\vec v_1\\ \Pi\to p\cdot\left(\begin{array}{c}n_x\\n_y\\n_z\end{array}\right)=p\cdot\vec n = 0 $$
now $r\in \Pi\Rightarrow (p_0+\lambda\vec v_0)\cdot \vec n = 0\ \ \forall\ \ \lambda$ giving
$$ \begin{cases} p_0\cdot\vec n = 0\\ \vec v_0\cdot \vec n = 0 \end{cases} $$
and finally $r_1\ \ |\ |\ \ \Pi\Rightarrow \vec v_1\cdot \vec n = 0$ hence the sought vector $\vec n$ is found by solving
$$ \begin{cases} p_0\cdot\vec n = 0\\ \vec v_0\cdot \vec n = 0\\ \vec v_1\cdot \vec n = 0\\ ||\vec n|| = 1 \end{cases} $$
or
$$ \left\{ \begin{array}{rcl} n_x+n_y& =& 0 \\ n_x+2 n_y+n_z& = & 0 \\ 2 n_x+n_y+n_z& = & 0 \\ n_x^2+n_y^2+n_z^2& =& 1 \\ \end{array} \right. $$
this system of equations does not have solution so I suggest to relax the plane definition from $p\cdot\vec n = 0\ $ to $(p-p^*)\cdot\vec n = 0$