= Solution
\b[The antisymmetric part need not be a <tensor>.] Its contraction with every symmetric test is zero, so the premise imposes no transformation law on it. This is the <blindness of symmetric contraction tests to antisymmetric arrays>, now with a third free index.
For a concrete example, prescribe the same numerical array in every orthonormal frame, with only
$$
d^a_{121}=1,\qquad d^a_{211}=-1
$$
nonzero, and take $d^s=0$. Every contraction $d^a_{ijk}s_{ij}$ vanishes, and the zero result is a genuine <vector> in every frame. But under $R=\operatorname{diag}(-1,-1,1)$ the rank-three <tensor> transformation law would send the $121$ component to $-1$, whereas our framewise prescription leaves it equal to $1$. Thus the array is not a <tensor> despite satisfying every stipulated contraction test. An antisymmetric part could be a <tensor> if an additional transformation law were supplied; it is simply not forced to be one.
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