Let $d_1:\begin{cases} 2x+y-z=1 \\ x-z=2\end{cases}$ and $d_2:\begin{cases} x-y+2z=1 \\ x-y=2\end{cases}$ be two lines in $\mathbb{R}^3$.
I have to find the plane $\pi_1$ that contains $d_1$ and is parallel to $d_2$ and the plane $\pi_2$ that is perpendicular on $\pi_1$ and contains $d_2$.
Here's what I did. The direction of $d_1$ is given by the cross product of the normal vectors of the two planes whose intersection it is. So, the direction of $d_1$ is the vector $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -1 \\ 1 & 0 & -1 \end{vmatrix}=(-1, 1, -1)$. Similarly, the direction of $d_2$ is $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -1 & 2 \\ 1 & -1 & 0 \end{vmatrix}=2(1, 1, 0)$, so we may consider the direction to be $(1, 1, 0)$ too.\
Now, $(1, -2, -1)\in d_1$, so $\pi_1:\begin{vmatrix} x-1 & -1 & 1 \\ y+2 & 1 & 1 \\ z+1 & -1 & 0\end{vmatrix}=0$. We get that $\pi_1 : x-y-2z=5$.
Since $\pi_2 \perp \pi_1$, their normal vectors must be perpendicular. We may take the normal vector of $\pi_2$ to be $(1, 1, 0)$ and now since $(2, 0, -\frac{1}{2})\in d_2$ we get that $\pi_2: 1(x-2)+1(y-0)+0(z+\frac{1}{2})=0$, which means that $\pi_2 :x+y=2$.
I think that this is correct, but one of the subsequent questions implied that $d_1$ and $\pi_2$ intersect and this is false with the relations I have. Did I do anything wrong?
You found the equation of $\pi_1$ correctly but for $\pi_2$, please note that it is perpendicular to $\pi_1$ but also contains line $d_2$.
Normal vector to $\pi_1$ is $(1, -1, -2)$. Also, direction vector of line $d_2$ is $(1, 1, 0)$. Normal vector to $\pi_2$ must be perpendicular to both vectors.
So the normal vector to $\pi_2$ can be obtained using cross product of $(1, -1, -2)$ and $(1, 1, 0)$ which gives us direction $(1, -1, 1)$.
As point $(2, 0, -\dfrac{1}{2})$ is on line $d_2$ (and so also on the plane $\pi_2$), equation of $\pi_2$ is $2x-2y+2z = 3$.