Problem that I need to solve is:

Let $\displaystyle a$ be line parallel to plane $\displaystyle \beta $ and let $\displaystyle \alpha $ be some plane that contains line $\displaystyle a$.

Then it's $\displaystyle \alpha \parallel \beta $ or intersection of these two planes is line parallel to $\displaystyle a$.

This is solution from book:

If $\displaystyle a \subset \beta $ then there are two cases (look at attached picture from book).

1) Planes $\displaystyle \alpha $ and $\displaystyle \beta $ have common only line $\displaystyle a$ so then is $\displaystyle \alpha \cap \beta = \{ a\} $, so $\displaystyle \alpha \cap \beta \parallel a$.

2) If there is at least one more point not belonging to line $\displaystyle a$ which is common to planes $\displaystyle \alpha $ and $\displaystyle \alpha '$, then is $\displaystyle \alpha = \alpha '$ so then is $\displaystyle \alpha \parallel \beta $. If it's $\displaystyle a \cap \beta = \emptyset $, then it's either $\displaystyle \alpha \parallel \beta $, that is $\displaystyle \alpha \cap \beta = \emptyset $, or it's $\displaystyle \alpha \cap \beta = a'$. All common points of planes $\displaystyle \alpha '$ and $\displaystyle \beta $ belongs to line $\displaystyle a'$, so since it's $\displaystyle \alpha \cap \beta = \emptyset $ then lines $\displaystyle a$ and $\displaystyle a'$ are not intersecting which means that they are parallel.

Solution from book is quite confusing to me.

Number 1) case is understable, nothing there to confuse.

Number 2) is not understable to me. First I will explain red text. Red text is reffering to case when $\displaystyle a \subset \beta $ and rest of text is reffered to case when $\displaystyle a \cap \beta = \emptyset $ (I think because author didn't specify that clearly). I don't understand why author introduced plane $\displaystyle \alpha '$ at all. I don't understand red text at all. Isn't solution just to say that if planes $\displaystyle \alpha $ and $\displaystyle \beta $ have one more common point not belonging to line $\displaystyle a$ then it's $\displaystyle \alpha \parallel \beta $?

The rest of text in 2) is quite confusing me.

My solution to when is $\displaystyle a \cap \beta = \emptyset $ is that if $\displaystyle \alpha \cap \beta = \emptyset $ then is $\displaystyle \alpha \parallel \beta $. If is $\displaystyle \alpha \cap \beta = a'$ then it must be $\displaystyle a\parallel a'$ because if $\displaystyle a$ isn't parallel to $\displaystyle a'$ then it would intersect plane $\displaystyle \beta $ which is contradiction. Also $\displaystyle a'$ can't intersect $\displaystyle a$ because then it would have only one common point with plane $\displaystyle \beta $ which is contradictory to $\displaystyle \alpha \cap \beta = a'$.

Can someone look at solution from book and my solution and give me explanation, comment or something that would clarify me that?:confused: