finding absolute value of sum of three non zero vectors

Question

If $\vec{u},\vec{v},\vec{w}$ are three non zero vectors and $\vec{u}\times \vec{v} = \vec{v}\times \vec{w} = \vec{w}\times \vec{u}.$ then $|\vec{u}+\vec{v}+\vec{w}|$ is

Attempt: using $\vec{u}\times \vec{v} = \vec{v}\times \vec{w}$ so $\vec{u}\times \vec{v} +\vec{w}\times \vec{v} = 0$ so $(\vec{u}+\vec{w})\times \vec{v}=0$

so $\vec{u}+\vec{w} = \lambda \vec{v}$ same way $\vec{u}+\vec{w} = \mu \vec{u}$ and $\vec{u}+\vec{v} = \gamma \vec{w}$

so $2(\vec{u}+\vec{v}+\vec{w}) = \lambda {v}+\mu \vec{u}+\gamma \vec{w}$

wan,t be able to go further,could some help me

Answer 1

We have $$\vec u+\vec v +\vec w = (\lambda+1)\vec v = (\mu+1)\vec u = (\nu+1)\vec w$$ One possibility is that the three vectors are all parallel, in which case $|\vec u+\vec v +\vec w|$ is a non-negative multiple of the magnitude of any one of them. If they are not parallel, then we have $\lambda=\mu=\nu=-1$ so $|\vec u+\vec v +\vec w|=0$. (In this case the vectors form the sides of a triangle; the given condition is just a restatement of the sine law for triangles.)