What does 'forms a right-handed set' mean? In a question I am reading, the following question appears.

What if $\vec{A},\vec{B}$, and $\vec{C}$ are mutually perpendicular and form a right-handed set?

What exactly does "form a right-handed set" mean? That $\hat{A}\times\hat{B}=\hat{C}$? Or does it mean the stronger equality $\vec{A}\times \vec{B}=\vec{C}$?
 A: There are several equivalent definitions, but here is one that is often convenient:
The ordered triplet $(\vec A,\vec B,\vec C)$ of vectors in $\mathbb R^3$ is right called right-handed if $\det(\vec A,\vec B,\vec C)>0$, where $(\vec A,\vec B,\vec C)$ is the square matrix formed by taking the three vectors as columns.
(Note that the determinant is necessarily nonzero since you assumed the vectors to be orthogonal and (hopefully also) nonzero.)
Similarly the triplet is left-handed if the determinant is negative.
The triplet is linearly dependent if the determinant is zero.
This generalizes to any dimension: we can easily define what it means for a $n$-tuplet of vectors in $\mathbb R^n$ to be right-handed.
If you multiply each vector in the multiplet by the same invertible square matrix, handedness is preserved if the matrix has positive determinant and reversed if negative.
Some intuitive properties are easy to observe from this point of view:
If you interchange any two vectors in the multiplet, the handedness is reversed.
If you permute the vectors cyclically, handedness is preserved.
A: We are accustomed to embed our  geometrical figures into the world around us, and in this environment the idea of "righhandedness" makes sense. But mathematically relevant is only "samehandedness", as follows:
Call two triples $(a_0,b_0,c_0)$, $\>(a_1,b_1,c_1)$ of linearly independent vectors in ${\mathbb R}^3$ (or any three-dimensional real vector space)  equivalent if the first triple can be continuously deformed into the second, such that at all times $t\in[0,1]$ the three vectors $a(t)$, $b(t)$, $c(t)$ are linearly independent. It turns out that there are exactly two equivalence classes, and that two triples are equivalent (or equally oriented) iff the matrix representing the one in terms of the other has positive determinant.
Usually one considers the standard basis of ${\mathbb R}^3$ as positively oriented and draws these three vectors in such a way that they seem "righthanded". Now for the cross product: Two linearly independent vectors $a$ and $b$ determine a two-dimensional plane whose orthogonal complement is a one dimensional vector space $L$, spanned by some vector $n\ne0$. Exactly one of $n$ and $-n$ complements $(a,b)$ to a positively oriented triple, and it is this choice of normal that is used in the definition of $a\times b$.
