Let $P\subset \mathbb{R}^3$ be a plane through the origin and $N$ be a unit normal to $P$. For $x,y \in P$, set $x\land y=(x \times y)\cdot N$.
Then for any three vectors $x,y,z \in P$, we have
$$(x\land y)z + (y\land z)x + (z\land x)y=0.$$
I can only think of changing the wedge products to determinants, and expand the left hand side to see if it becomes $0$, but this is very tedious and I'm not sure it works since I need to use the condition that $(x \times y)$ is parallel to $N$. But I can't think of any simpler ways. How can I show this? I would greatly appreciate any help.
Three vectors $x,y,z$ in a plane are necessarily linearly dependent. Without loss of generality, we may assume $z$ is a linear combination of $x$ and $y$, i.e. $z=\lambda x+\mu y$. Then $$(x\wedge y)z+(y\wedge z)x+(z\wedge x)y=\lambda(x\wedge y)x+\mu(x\wedge y)y+\lambda (y\wedge x)x+\mu(y\wedge y)x+\lambda(x\wedge x)y+\mu(y\wedge x)y.$$ Using skew-symmetry of the wedge product, this expression is seen to be zero.