= Solution
A <Poisson structure> is a smooth antisymmetric <matrix> field $\omega^{ab}$ defining a bracket
$$
\{f,g\}=\sum_{a,b}\omega^{ab}\partial_af\,\partial_bg
$$
that obeys the <Jacobi identity>. Bilinearity, antisymmetry and the product rule follow from this formula; Jacobi is the remaining condition. Applying Jacobi to the coordinate functions gives
$$
\boxed{\sum_d\left(\omega^{dc}\partial_d\omega^{ab}
+\omega^{db}\partial_d\omega^{ca}
+\omega^{da}\partial_d\omega^{bc}\right)=0.}
$$
These are also sufficient: expanding Jacobi for arbitrary functions cancels all second-derivative terms by antisymmetry, leaving these coefficients multiplying their first derivatives.
For $\omega^{ab}=\epsilon^{abc}x^c$, the nonconstant function $f=\sum_a(x^a)^2$ is a <Casimir function of a Poisson manifold>, because
$$
\{f,x^a\}=2\sum_{b,c}x^b\epsilon^{bac}x^c=0
$$
by symmetry of $x^bx^c$. This is the usual three-dimensional <Lie-Poisson bracket>.
Since the <matrix> $M$ is symmetric, $\partial_dH=\sum_bM^{db}x^b$. The <Hamilton equations> are $\dot x^a=\{x^a,H\}$, and hence
$$
\dot x^a=\sum_{b,c,d}\epsilon^{adc}M^{db}x^bx^c.
$$
One valid choice of coefficients is $Q^{abc}=\sum_d\epsilon^{adc}M^{db}$. If symmetric coefficients in $b,c$ are preferred, the same equations use
$$
\boxed{Q^{abc}=\frac12\sum_d
\left(\epsilon^{adc}M^{db}+\epsilon^{adb}M^{dc}\right).}
$$
In vector notation the evolution is $(Mx)\times x$, consistent with the bracket convention and with preservation of $|x|^2$.
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