= Solution
Use $\mathbf w=-\mathbf u-\mathbf v$ and bilinearity and antisymmetry of the <cross product>:
$$
\mathbf v\times\mathbf w=-\mathbf v\times\mathbf u-\mathbf v\times\mathbf v=\mathbf u\times\mathbf v,
\qquad
\mathbf w\times\mathbf u=-\mathbf u\times\mathbf u-\mathbf v\times\mathbf u=\mathbf u\times\mathbf v.
$$
Consequently,
$$
\boxed{\mathbf u\times\mathbf v=\mathbf v\times\mathbf w=\mathbf w\times\mathbf u,\qquad
\operatorname{span}\{\mathbf u,\mathbf v,\mathbf w\}=\operatorname{span}\{\mathbf u,\mathbf v\}.}
$$
If $\mathbf u,\mathbf v$ are <linearly independent>, this <linear span> is the <plane> through the origin with <normal vector> $\mathbf u\times\mathbf v$. If they are dependent, all three nonzero <vectors> lie on one <line> through the origin, and their <linear span> has dimension one. For example, $\mathbf u=\mathbf v=(1,0,0)$ and $\mathbf w=(-2,0,0)$ obey the hypothesis but span only a <line>. Nonzero <vectors> alone do not force a two-dimensional <linear span>.
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