= Solution
For a <Cartesian second-rank tensor>, write $S_{ij}=(A_{ij}+A_{ji})/2$ and $W_{ij}=(A_{ij}-A_{ji})/2$. Under an orthogonal change of frame,
$$
A'_{ij}=R_{ik}R_{jl}A_{kl},\qquad
S'_{ij}=R_{ik}R_{jl}S_{kl},\qquad W'_{ij}=R_{ik}R_{jl}W_{kl},
$$
where the last two equalities follow by interchanging the dummy indices in $A'_{ji}$. Thus the symmetric and antisymmetric parts obey the same tensor transformation law. If $A=S+W$ with $S^T=S$, $W^T=-W$, adding or subtracting its transpose forces $S=(A+A^T)/2$ and $W=(A-A^T)/2$. This also proves uniqueness.
For the given matrix, the scalar part is $a=\operatorname{Tr}(A)/3=3$. The symmetric part is $S=\begin{pmatrix}1&3&2\\3&5&4\\2&4&3\end{pmatrix}$, so the <traceless second-rank tensor> is $B=S-3I$. The skew part is $W=\begin{pmatrix}0&-1&1\\1&0&2\\-1&-2&0\end{pmatrix}$. Comparing with the <cross product> matrix $[p]_\times=\begin{pmatrix}0&-p_3&p_2\\p_3&0&-p_1\\-p_2&p_1&0\end{pmatrix}$ gives
$$
\boxed{a=3,\qquad p=(-2,1,1)^T,\qquad B=\begin{pmatrix}-2&3&2\\3&2&4\\2&4&0\end{pmatrix}.}
$$
Then $A=aI+[p]_\times+B$ proves the decomposition for every vector. The vector dual to the antisymmetric tensor is an <axial vector> under reflections, and an ordinary vector under proper rotations.
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