= Solution
Use the symmetric <joint probability density function> of exchangeably labelled <eigenvalues>, normalized over all of $\mathbb R^3$. The <finite-rank projection kernel> has
$$
\int\widetilde K_3(x,x)\,dx
=\sum_{\ell=0}^2\lVert\phi_\ell\rVert_2^2=3.
$$
Successive applications of <Gaudin's lemma> therefore give
$$
\int_{\mathbb R^3}\det[\widetilde K_3(\lambda_i,\lambda_j)]_{i,j=1}^3
\,d\lambda_1d\lambda_2d\lambda_3
=1\cdot2\cdot3=6.
$$
It follows that
$$
\boxed{\widetilde c_3=\frac1{3!}=\frac16.}
$$
As a normalization check, the <weighted Vandermonde determinant> yields
$$
\det[\widetilde K_3(\lambda_i,\lambda_j)]
=\frac{\prod_{i<j}(\lambda_i-\lambda_j)^2e^{-\sum_i\lambda_i^2/2}}
{\kappa_0^2\kappa_1^2\kappa_2^2},
\qquad
\boxed{c_3=\frac1{12(2\pi)^{3/2}}.}
$$
If 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.
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