= Solution
Let $s=A_iA_i>0$, $q=A_iT_{ij}A_j$ and $\tau=T_{ii}$. Taking the <trace> of the proposed decomposition gives
$$
\tau=3\alpha+\beta s,
$$
since $A_iB_i=0$ and $C_{ii}=0$. Contracting twice with $A$ gives $q=\alpha s+\beta s^2$. Solving these two scalar equations yields
$$
\boxed{\alpha=\frac12\left(\tau-\frac qs\right),\qquad
\beta=\frac{3q-s\tau}{2s^2}.}
$$
A single contraction gives $T_{ij}A_j=(\alpha+\beta s)A_i+sB_i$, and $\alpha+\beta s=q/s$. Hence
$$
\boxed{B_i=\frac{T_{ij}A_j}{s}-\frac{qA_i}{s^2}.}
$$
Finally set
$$
\boxed{C_{ij}=T_{ij}-\alpha\delta_{ij}-\beta A_iA_j-B_iA_j-B_jA_i.}
$$
This is manifestly symmetric. The formulas give $A_iB_i=q/s-q/s=0$; substituting them into $C_{ij}A_j$ gives zero, and taking the <trace> gives $C_{ii}=0$. Since $C$ is symmetric, $A_iC_{ij}=0$ also holds.
A geometric form makes the <axis decomposition of a symmetric matrix> explicit. With the transverse <orthogonal projection> $P_{ij}=\delta_{ij}-A_iA_j/s$,
$$
B_i=\frac{P_{ij}T_{jk}A_k}{s},\qquad
C_{ij}=P_{ik}T_{kl}P_{lj}-\alpha P_{ij}.
$$
Here $\alpha$ is half the transverse <trace> and $C$ is its <symmetric traceless rank-two tensor> part. Every term was forced by contractions, so the decomposition is unique.
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