Volume of a paralellepiped The question I am asked to answer is the following: 
Let $\vec{a} = (1,1,1)$ and $\vec{b} = (1,1,-1)$. Find all vectors $\vec{x} = (x,y,z)$ in $\mathbb{R}^3$ such that the ordered vector triple $(\vec{a},\vec{b},\vec{x})$ represents the right hand orientation of $\mathbb{R}^3$ and such that the paralellepiped spanned by the vectors has volume $2$. Which object in $\mathbb{R}^3$ does the collection of all such vectors form?
My attempt:
I know that the volume of a paralellepiped defined by vectors $\vec{a}$, $\vec{b}$ and $\vec{x}$ is equal to the magnitude of the mixed product, which should equal $2$ in this case. However, that gives me only one equation; when I solve the mixed product, I get $2x-2y = 2$.  
I think the problem is that I really don’t understand what the question is asking. What is meant by right hand orientation of $\mathbb{R}^3$? I checked in my book and notes and can’t find any formal explanation. Also, I have no idea which object in  $\mathbb{R}^3$ is formed by the collections of all such vectors $\vec{x}$.
Thank you for any help! 
 A: Here are some hints.
You know that the volume is the magnitude of the mixed product, so the that the mixed product is $\pm 2,$ but the problem says the the volume is equal to the mixed product itself, so that the mixed product is $2.$  Now, I suppose you know that if two of the vectors in the mixed product are swapped, the sign changes. The "right-hand orientation" means that you should compute $\vec a \cdot (\vec b \times \vec x),$ which it appears to me you have done.  
For the equation you got, $2x-2y=2,$ this says that $y = x-1$ so the solutions are all vectors of the form $(x,x-1,z).$  These are all the solutions to the equation $y=x-1.$
I'm sure you know that this is a plane.  If it seems weird that the solution set is so big, remember that the volume of a parallelepiped is the base time the height. The given vectors $\vec a, \vec b$ lie in the plane $x=y$ and the solution is telling you that the height has to be one.
But there are two planes at distance $1$ from $x=y.$  Why isn't the answer $y=x+1?$  This is where the business of the right-hand orientation and the magnitude comes in.  With those specifications, there is only one plane in the solution.  If the right-hand orientation were not specified, it woould be perfectly legitimate to solve $\vec a \cdot (\vec x \times \vec b) = 2,$ and if I'm not mistaken that would give tou the other plane.  Try it and see!
In answer to your question, no you don't have to solve for the $z$ component.  The work you have already done shows that any value of $z$ will work.  Look back at your work.  See how $z$ drops out?  It doesn't matter what it is.  
A: If we want that the three vectors $\vec a$, $\vec b$, and $\vec x$ form a positively oriented triple then we have to make sure that triple vector product $$[\vec a,\vec b,\vec x]=(\vec a\times\vec b)\cdot\vec x$$
is positive. If this is the case then the positive value of this triple product is equal to the volume ${\rm vol}(P)$ of the parallelepiped spanned by $\vec a$, $\vec b$, and $\vec x$. Since we want the three vectors be positively oriented and this volume to be $=2$ this means that $\vec x$ has to satisfy the condition
$$ (\vec a\times\vec b)\cdot\vec x=(-2,2,0)\cdot(x,y,z)=2\ ,$$
or 
$$-2x+2y=2\ ,$$
which is the equation of a plane in ${\mathbb R}^3$.
