 Solution

Use the convention ${\Theta }^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{e}^c\wedge e^d$ and $R_{ab}=R^c{}_{acb}$. Each independent two-form has only one coordinate-plane wedge, so the Ricci tensor is diagonal. Contracting the six coefficients givesBut $2B+F=2(1+q/2)+(1-q)=3$, soIn coordinate-independent form, $R_{\mu \nu }=-3g_{\mu \nu }$. Thus the vacuum Einstein field equations $G_{\mu \nu }+\Lambda g_{\mu \nu }=0$ hold withThe curvature radius is one in the metric's normalization. The parameter $\alpha$ changes the individual curvature 2-forms but cancels from their Ricci contraction. For $\alpha =0$ this is Anti-de Sitter spacetime; for nonzero $\alpha$ it is the planar Einstein metric with cubic radial function, with a nontrivial Weyl tensor rather than universally constant sectional curvature. The result is local on regular coordinate patches; a zero of $F$ invalidates this particular static coframe, not the tensor equation in a suitable regular extension.