= Solution
Here <isotropic tensors> are invariant under proper <rotation matrices>, $R^TR=I$ and $\det R=1$. The <isotropic second-rank tensor> and <isotropic third-rank tensor> forms in three dimensions are
$$
\boxed{T_{ij}=A\delta_{ij},\qquad T_{ijk}=B\epsilon_{ijk},}
$$
with scalar coefficients $A,B$. One can see completeness directly. A half-turn about each coordinate axis kills every off-diagonal component of a rank-two <tensor>, and quarter-turns make its three diagonal components equal. For rank three, the half-turns require that each coordinate label occur an odd number of times in a nonzero component. Thus only permutations of $(1,2,3)$ survive. Quarter-turns interchange two labels while reversing one sign, forcing the surviving components to alternate with the <parity of a permutation>. This gives a multiple of the <Levi-Civita symbol>.
The <tensor> transformation law verifies invariance of these forms:
$$
T'_{ij}=AR_{ip}R_{jq}\delta_{pq}=A\delta_{ij},\qquad
T'_{ijk}=BR_{ip}R_{jq}R_{kr}\epsilon_{pqr}=B(\det R)\epsilon_{ijk}=B\epsilon_{ijk}.
$$
The first equality uses <orthogonality> and the second uses the <determinant> formula. If reflections were also required while using the ordinary <tensor> transformation law, the rank-three coefficient would have to vanish. The proper-rotation convention is the one under which $B\epsilon_{ijk}$ is an <isotropic tensor>.
The ball is unchanged by any <rotation matrix>, and rotating coordinates has <Jacobian determinant> one. Thus changing variables $\mathbf y=R\mathbf x$ gives
$$
R_{i_1j_1}\cdots R_{i_nj_n}\int_Vx_{j_1}\cdots x_{j_n}\,dV
=\int_Vy_{i_1}\cdots y_{i_n}\,dV=T_{i_1\cdots i_n}.
$$
These moments are therefore <isotropic tensor integrals>. For the <Cartesian moments of a three-dimensional ball>, the second moment is $\alpha\delta_{ij}$. Taking its <tensor contraction> gives
$$
3\alpha=\int_Vr^2\,dV=4\pi\int_0^a r^4\,dr=\frac{4\pi a^5}{5},\qquad
\boxed{\alpha=\frac{4\pi a^5}{15}.}
$$
The third moment vanishes: $\mathbf x\mapsto-\mathbf x$ preserves the ball and changes the sign of its integrand. Equivalently, its symmetry under index interchange excludes a nonzero multiple of the antisymmetric <Levi-Civita symbol>.
The fourth moment is symmetric under every <permutation> of its four indices. The general rank-four <isotropic tensor> form consequently has equal coefficients, yielding
$$
T_{ijkl}=\beta(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}).
$$
For example, comparing $T_{1122},T_{1212},T_{1221}$ shows that the three coefficients agree. Contracting $i=j$ and $k=l$ gives
$$
15\beta=\int_V r^4\,dV=4\pi\int_0^a r^6\,dr=\frac{4\pi a^7}{7},\qquad
\boxed{\beta=\frac{4\pi a^7}{105}.}
$$
Finally the <vector triple product identity> gives $\mathbf x\times(\boldsymbol\Omega\times\mathbf x)=r^2\boldsymbol\Omega-(\boldsymbol\Omega\cdot\mathbf x)\mathbf x$. The first term integrates to $3\alpha\boldsymbol\Omega$ and the second to $\alpha\boldsymbol\Omega$, using the second moment. Hence
$$
\boxed{\int_V\mathbf x\times(\boldsymbol\Omega\times\mathbf x)\,dV=\frac{8\pi a^5}{15}\boldsymbol\Omega.}
$$
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