Set equality proof 


I know that for this proof I need to show that $A$ $\subseteq$ $B$ and $B$ $\subseteq$ $A$. Starting with $A$ $\subseteq$ $B$, I started by setting the equations equal to each other and solving for $x$.  
$2x$ - $y$ + $7z$ = $x$ - $y$ + $5z$ gives,  $x$ = $-2z$. Plugging $x$ back into one of the equations, you get $(-2z)$ -$y$ + $7z$ = $0$. Simplifying, I get $y$ = $3z$. So, for all ($x$, $y$, $z$) $\in$ $A$, ($x$, $y$, $z$) = ($-2z$, $3z$, $z$). That is, ($x$, $y$, $z$) = ($-2c$, $3c$, $c$) since we took ($x$, $y$, $z$) to be arbitrary. So ($x$, $y$, $z$) $\in$ $B$ and $A$ $\subseteq$ $B$. 
Now, I'm having trouble going in the other direction and showing that $B$ $\subseteq$ $A$. I have the following thus far: Assume ($x$, $y$, $z$) $\in$ $B$. Thus, ($x$, $y$, $z$) = ($-2c$, $3c$, $c$). I'm not sure how to get from here to showing that $B$ $\subseteq$ $A$.
 A: Take any $(x,y,z)\in B$, then $x= -2c$, $y=3c$ and $z=c$, so $$x-y+5z = -2c-3c+5c =0$$ and $$2x-y+7z = -4c-3c +7c =0$$ so $(x,y,z)\in A$ and thus $B\subseteq A$.
A: $A$ is a line: it is an intersection of two non-parallel planes.
$B$ is a line, explicitly parameterized by $c$.
Therefore:
To show $B \subset A$, all you need is two points from $B$ that lie in $A$. Direct inspection shows that the origin ($c=0$) does.  This is one point.
To get another, take, for example, $c = 1$, so the corresponding point in $B$ is:
$$
(x, y, z) = (-2, 3, 1). 
$$
Plug it into the two equations that define $A$ and see that they are satisfied.
A: Try not to get too caught up in set concepts.
$A \subset B$ means $x-y+5z = 0; 2x -y +7z = 0 \implies \exists c; x=-2c;y=3c; z = c$.
And $B \subset A$ means if we let $z=c; y=3c; x=-2c \implies  x-y+5z = 0; 2x -y +7z = 0$
And $A = B$ means precisely $x-y+5z = 0; 2x -y +7z = 0 \iff \exists c; x=-2c;y=3c; z = c$
...
To be honest $B \subset A$ is the easier way:
If $x = -2c; y = 3c; z = c$ then $x - y + 5z = (-2c) -3c + 5c =-5c + 5c= 0$ and $2(-2c)-3c +7c = -4c-3c + 7 = -7c + 7c=0$ and we are done for that direction.
That was it.
$A\subset B$ is a matter of showing:
$x - y + 5z = 0; 2x -y + 7z = 0$ is two equations with three unknowns.  If we solve in terms of $z$ we get (one way or another[1]) $y = 3z$ and $x = -2z$ and that's it.  Set $z=c$ and we are done.
[1] One way:
$y = x+ 5z = 2x + 7z$
$2x - x = 5z - 7z$
$x = -2z$
$y = -2z + 5z = 2(-2z) + 7z = 3z$.
A: Set $A$: Intersection of $2$ planes passing through the origin determines a straight line passing through the origin.
Ansatz:
$\vec r = t(a,b,c)$, $t \in \mathbb{R}$.
Determine the direction vector $(a,b,c)$.
1) $ta -tb +t5c=0$;
2) $2ta-tb+7tc=0$;
1)$a-b+5c=0;$
2)$2a-b+7c=0$;
$a+2c=0$; $a =-2c;$
$b= -2c+5c=3c$;
$(a,b,c)= (-2c,3c,c)$.
Hence:  $\vec r = t (-2,3,1).$
Finally :
$A=$
{$(x,y,z)=t(-2,3,1), t \in \mathbb{R}$}
$=B.$
