= Solution
In orthonormal <Cartesian coordinates>, a rank-$n$ <tensor> has components transforming under a change of axes $R$ as
$$
\boxed{T'_{i_1\cdots i_n}=R_{i_1j_1}\cdots R_{i_nj_n}T_{j_1\cdots j_n},\qquad R^TR=I.}
$$
Repeated indices are summed. This is the <tensor component transformation law>; for rank one it is precisely the transformation law of a <vector>.
Use the <vector triple product> to write the double <cross product> as
$$
W=u\times(\Omega\times u)=(u\cdot u)\Omega-(u\cdot\Omega)u.
$$
Both <dot products> are invariant under an orthogonal change of axes, while $u'=Ru$ and $\Omega'=R\Omega$. Therefore
$$
W'=(u\cdot u)R\Omega-(u\cdot\Omega)Ru=RW.
$$
Hence \b[$W$ is a vector]. This argument also avoids the sign issue for an individual <cross product> under an orientation-reversing orthogonal transformation: the double product has two such factors, whose signs cancel.
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