A question on parallelism of affine planes

Question

Disclaimer: I am not sure how much of this is common terminology/ notation, so please bear with me if anything is unclear, I will supply the needed information.
By affine plane we mean an affine subspace of dimension $2$.
So, I learnt that if we have an affine space $\mathcal{A}$ and two affine subspaces $\mathcal{A}_1$ and $\mathcal{A}_2$, then we say that $\mathcal{A}_1$ and $\mathcal{A}_2$ are parallel if $\operatorname{dir}(\mathcal{A}_1)\subset \operatorname{dir}(\mathcal{A}_2)$ or $\operatorname{dir}(\mathcal{A}_2)\subset \operatorname{dir}(\mathcal{A}_1)$, where by $\operatorname{dir}(\mathcal{A}_i)$ I denoted the direction of $\mathcal{A}_i$.
Let's consider this problem now: Let $\pi \subset \mathbb{C}^3$, $\pi:\sqrt{2} x +(1-\sqrt 3)y -z -1 =0$ and $P:=(1,0,2)$. Find the equation of the plane $\alpha$ such that $P\in \alpha$ and $\alpha \parallel \pi$.
My instructor said that $\alpha \parallel \pi \iff \operatorname{dir}(\mathcal{A}_1)=\operatorname{dir}(\mathcal{A}_2)$. This isn't consistent with the definition from above. I think that what he meant was that this equivalence holds because we require $P\in \pi$ and by the fifth postulate I think that we get that equivalence. Am I right?
Furthermore, he went on to say that $\operatorname{dir}(\alpha):\sqrt 2 x+(1-\sqrt 3)y-z=0$. Why? Does this hold because $\operatorname{dir}(\alpha)$ is a vector subspace of $\mathbb{C}^3$?

Answer 1

Then direction of an affine plane in $K^3$ with equation $ax+by+cz+d=0$, is the vector subspace with equation $ax+by+bz=0$, and the affine plane with the same direction that passes though the point $P_0(x_0,y_0z_0)$ is defined by the equation $$ax+by+bz=ax_0+by_0+cz_0.$$

$\DeclareMathOperator{\dir}{dir}$ Also, $\alpha\parallel\pi$ indeed means $\dir\mathcal A_1\subseteq\dir\mathcal A_2$, but as the spaces have the same dimension, it is equivalent to $\dir\mathcal A_1=\dir\mathcal A_2$.