Constructing the vector product having in mind that its norm should be an area

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I only saw constructions where we posit the different relations such as $\hat\imath\times\hat\jmath=\hat k$ etc, but I want to do is this.

I want a vector yielding operation $\vec w=\vec u\times\vec v$ that would give me:

  1. $\vec w\cdot\vec u=0$
  2. $\vec w\cdot\vec v=0$
  3. $|\vec w|=|\vec u||\vec v|\sin\left(\angle(\vec u,\vec v)\right)$

So if $\vec w=(a,b,c)$, $\vec u=(d,e,f)$, and $\vec v=(g, h, i)$, I have:

  1. $ad+be+cf=0$
  2. $ah+bh+ci=0$
  3. $a^2+b^2+c^2=(d^2+e^2+f^2)(g^2+h^2+i^2)-(dg+eh+fi)^2$

I squared $(3)$ and got the sine from $\vec u\cdot\vec v$.

What next, though?

Frankly, I don't like to think that we have the norm of the vector product giving us the area of the parallelogram described by both vectors be just a happy coincidence.

Edit:

I can expand $3$ to get $|\vec w|^2=(ei-hf)^2+(di-gf)^2+(dh-eg)^2$, which is literally the norm of the usual vector product, but I haven't at all used $1$ and $2$. It seems too arbitrary to just assume $(a,b,c)=(ei-hf,di-gf,dh-eg)$, especially knowing that I have ordered them in that specific manner because of my prior knowledge of the $\vec w$'s coordinates.