I am trying to understand how $(1),(2) \text{ and } (3)$ are valid if the coordinate is a right-handed coordinate system. The definition of a right-handed coordinate system is a one such that $\overrightarrow{i} \times \overrightarrow{j} = \overrightarrow{k}$ is valid. To me, to prove that I must compute all possibilities to verify that result. Is there any other method?
$$\overrightarrow{i} \times \overrightarrow{j} = \overrightarrow{k} \qquad(1)\\ \overrightarrow{j} \times \overrightarrow{k} = \overrightarrow{i} \qquad(2)\\ \overrightarrow{k} \times \overrightarrow{i} = \overrightarrow{j} \qquad(3)$$
I want to very that $(1),(2) \text{ and } (3)$ is valid as long as it is a right-handed coordinate system. what I mean by possibilities is to try all the different ways that $x,y$ and $z$ can be. For example:
[https://en.wikipedia.org/wiki/Cross_product#Computing_the_cross_product]
$\vec i \times \vec j = \vec k$ in a right-handed orthonormal coordinate system. That is, a coordinate system where the three basis vectors are all orthogonal to each other and have length $1$. General coordinate systems may have $\vec i \times \vec j \ne \vec k$. They are called right-handed when $(\vec i \times \vec j) \cdot \vec k > 0$.
I will assume that you are only asking about orthonormal coordinate systems. One identity for the cross product is the "bac-cab" rule: $$\vec a \times (\vec b \times \vec c) = \vec b(\vec a\cdot\vec c) - \vec c(\vec a \cdot \vec b)$$
Since the coordinate system is orthonormal, if $\vec i \times \vec j = \vec k$, then $$\vec j \times \vec k = \vec j \times (\vec i \times \vec j) = \vec i (\vec j \cdot \vec j) - \vec j(\vec j \cdot \vec i) = 1\vec i - 0\vec j = \vec i$$ $$\vec k \times \vec i = \vec k \times (\vec j \times \vec k) = \vec j(\vec k \cdot \vec k) - \vec k (\vec k \cdot \vec j) = 1\vec j - 0\vec k = \vec j$$