= Solution
Split the array in its first two slots:
$$
d^s_{ijk}=\frac{d_{ijk}+d_{jik}}2,\qquad
d^a_{ijk}=\frac{d_{ijk}-d_{jik}}2.
$$
For every <symmetric second-rank tensor> $s$, antisymmetry gives $d^a_{ijk}s_{ij}=0$. Hence the stipulated <vector> equals $d^s_{ijk}s_{ij}$.
Let $R$ change the orthonormal frame, so $s'_{ab}=R_{ai}R_{bj}s_{ij}$. The <vector> transformation law gives
$$
d^{s\prime}_{abc}R_{ai}R_{bj}s_{ij}
=R_{ck}d^s_{ijk}s_{ij}
$$
for every symmetric test $s$. Both coefficient arrays in $i,j$ are symmetric, so equality against every symmetric <matrix> forces equality of the coefficients; one may test the individual diagonal entries and the symmetric off-diagonal <basis> <matrices>. Multiplying by $R_{ai}R_{bj}$ and using orthogonality yields
$$
\boxed{d^{s\prime}_{abc}=R_{ai}R_{bj}R_{ck}d^s_{ijk}.}
$$
This is exactly the rank-three <tensor> law. It is the <symmetric-test criterion for a third-rank Cartesian tensor>, a form of the <quotient theorem for Cartesian tensors>. Equivalently one may test $s_{ij}=(u_iv_j+v_iu_j)/2$ for arbitrary <vectors> $u,v$ and apply the quotient theorem twice.
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