= Solution
The <special orthogonal group> in three dimensions is
$$
SO(3)=\{A\in M_3(\mathbb R):A^TA=I,\ \det A=1\}.
$$
Differentiating $A(t)^TA(t)=I$ at the identity shows that its <Lie algebra> is the space of <skew-symmetric matrices>. A convenient basis is
$$
J_1=\begin{pmatrix}0&0&0\\0&0&-1\\0&1&0\end{pmatrix},
\quad
J_2=\begin{pmatrix}0&0&1\\0&0&0\\-1&0&0\end{pmatrix},
\quad
J_3=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix},
$$
with $[J_a,J_b]=\epsilon_{abc}J_c$.
The infinitesimal action of $\exp(tJ_a)$ on a point $x\in\mathbb R^3$ is $J_ax=e_a\times x$. It therefore generates the vector field
$$
\boxed{V_a=\epsilon_{abc}x_b\frac{\partial}{\partial x_c}}.
$$
Fundamental vector fields for this left action form an antihomomorphism with the stated convention, and direct differentiation gives
$$
\boxed{[V_a,V_b]=-\epsilon_{abc}V_c},
$$
so their span is closed under the <Lie bracket of vector fields>.
The brackets $\{x,y\}=z$, $\{y,z\}=x$, and $\{z,x\}=y$ define the <rotational Lie-Poisson structure on R3>. With the convention that a <Hamiltonian vector field> acts by $X_H(f)=\{f,H\}$,
$$
X_{x_a}(x_c)=\{x_c,x_a\}=\epsilon_{abc}x_b=V_a(x_c).
$$
Hence the required Hamiltonians are simply
$$
\boxed{H_a=x_a}.
$$
The quadratic function
$$
\boxed{F=x^2+y^2+z^2}
$$
satisfies $\{F,x_a\}=0$ for all $a$, so it is a <Casimir function of a Poisson manifold>. Its nonzero regular level sets $F=R^2$ are spheres. The Poisson tensor has rank two there and is tangent to each level set, so it inverts to a <symplectic form>; each sphere is a <symplectic leaf>. Rotations preserve both $F$ and the alternating tensor $\epsilon_{abc}$, hence preserve the restricted Poisson tensor and its inverse symplectic form. The $SO(3)$ action therefore restricts to a symplectic action on every sphere $S_R^2\subset\mathbb R^3$.
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