 Solution

Use the symmetric joint probability density function of exchangeably labelled eigenvalues, normalized over all of ${\mathbb{R}}^3$. The finite-rank projection kernel hasSuccessive applications of Gaudin's lemma therefore giveIt follows thatAs a normalization check, the weighted Vandermonde determinant yieldsIf instead one uses ordered eigenvalues on ${\lambda }_1<{\lambda }_2<{\lambda }_3$, the normalized density on that chamber is six times the labelled density. In that convention the determinant prefactor on the chamber is one.