Doubts about Chern-Simons state as a solution of the Hamiltonian constraint in quantum gravity

Question

I've been doing some work with both Baez's *Knots, gauge fields and gravity* (1) and Gambini, Pullin's *Loops, knots, gauge Theories and quantum gravity* (2), lately.

I have basically two problems: I understand that, in the ADM formalism, the Lagrangian density for the cosmological term of Einstein equation is given by

$$L= q\Lambda \underline{N},$$
where $q$ is the determinant of the 3-metric, $\Lambda$ is the cosmological constant, and $\underline{N}$ is $q^\frac{-1}{2}$N (the lapse function). Also, that
$$\tilde{E^i_a}=q^{\frac{1}{2}}E^i_a$$
are the densitized triads of the Ashtekar formalism. However, I don't get why $q$ can be given by the expression (7.53) from (2):
$$q=\frac{1}{6}\underline{\epsilon_{abc}}\epsilon^{ijk}\tilde{E^a_i}\tilde{E^b_j}\tilde{E^c_K}.$$
Is there a way to obtain such expression?

The second problem is: after promoting the Ashtekar variables to operators ( $\hat{A^a_i}$ and $\hat{E^a_i}=\frac{\delta}{\delta A^a_i}$ ), it's can be shown, for the Chern-Simons state

$$\psi_{\Lambda}= e^{-\frac{6}{\Lambda}S_{CS}},$$
with $S_{CS}$ being the Chern-Simons action
$$S_{CS}= \int_{\Sigma} tr \,(A\wedge dA +\frac{2}{3}A\wedge A\wedge A),$$
that
$$\begin{eqnarray*} \frac{\delta}{\delta A^i_a}\psi_\Lambda = \frac{3}{\Lambda}\overline{\epsilon^{abc}}F^i_{bc}\psi_\Lambda \\ \underline{\epsilon_{abc}}\frac{\delta}{\delta A^i_a}\psi_\Lambda = \frac{6}{\Lambda}F^i_{bc}\psi_\Lambda, \end{eqnarray*}$$
which comes from expressions (7.70) and (7.71) from (2).

My problem is with the second line. Am I supposed to take 

$$\underline{\epsilon_{abc}}\overline{\epsilon^{abc}} = 2~?$$  Why would that be true?

Sorry for the lengthy post. I'd be glad if someone could help me with these.