I am wondering why the cross product of two vectors in $\mathbb{R}^3$ would get the Normal Vector of the plane generated by them? I know it is the definition, but I am still wondering why we can get it just by calculating their cross product.
Is there any proof?
This is cross product of two vectors in $\mathbb{R}^3$ :
$$(a,b,c) \times (d,e,f) = (bf-ce, cd-af, ae-bd)$$
When you believe that the scalar product of two non-zero vectors is 0 exactly when they are perpendicular, then its easy. You just have to evaluate:
$\mathbf{a}\cdot(\mathbf{b} \times \mathbf{c})$
for either $\mathbf{a} = \mathbf{b}$ or $\mathbf{a} = \mathbf{c}$. You can do this using the coordinate expansions you mentioned above or using algebraic properties such as the triplet product expansion of the cross product/scalar product together with the anti-commutativity of the cross product $\mathbf{a} \times \mathbf{b} = - \mathbf{b} \times \mathbf{a}$ (and here setting $\mathbf{a} = \mathbf{b}$, this must then be the $\mathbf{0}$ vector).