The existence of a finite dimensional Lie algebra with a given symmetric invariant metric The question is motivated by a more broad perspective in another MO post and here, but here we would like to understand a specific case (our question potentially connects to / is motivated b Quantum Gravity).
Question: We would like to know whether there exists a finite (here six) dimensional Lie algebra which provides a symmetric invariant bilinear non-degenerate metric?
$$
\omega_{ab}\equiv
{\begin{pmatrix} 
 \Omega_{h_i, h_j} & a_{i}\;\mathbb{I}\\
  a_{i}\;\mathbb{I} & 0
\end{pmatrix}}
$$
With $\Omega_{h_i, h_j}\equiv
{\begin{pmatrix} 
2b_1 & c_{12}& c_{13} \\
c_{12} & 2b_2 & c_{23} \\
c_{13} & c_{23} & 2b_3 
\end{pmatrix}}$, and
$a_{i}\mathbb{I}\equiv
{\begin{pmatrix} 
a_1 &0 &0 \\
0 &a_2 &0 \\
0& 0& a_3 
\end{pmatrix}}$
(Here $a_1,a_2,a_3$ all are positive numbers. $b_1,b_2,b_3,c_{12},c_{13},c_{23}$ are either positive numbers or zeros.) if yes, what is this Lie algebra? The commutation rules? Is it semi-simple Lie algebra or non-semi-simple? nilpotent or not, solvable or non-solvable Lie algebra?

extra INFO:
NOTE: A specific case we had worked out (but which unfortunately whose Lie algebra does not provide the desired $\omega_{ab}$) is the doubled extension of $\mathcal{G}$ by $\mathcal{h}$. The full vector space is $\mathcal{G}\oplus \mathcal{h}\oplus \mathcal{h}^*$. Here the subalgebra is spanned by the Abelian extension of $\mathcal{G}$ by $\mathcal{h}^*$: i.e. $\mathcal{G}\oplus \mathcal{h}^*$. The full algebra is the semi-direct product of $\mathcal{h}$ by this Abelian extension, $\mathcal{G}\oplus \mathcal{h}^*$. This is the doubled extension:  $\mathcal{G}\oplus \mathcal{h}\oplus \mathcal{h}^*$. Also $\mathcal{h}^*$ is the dual to $\mathcal{h}$. This Lie algebra is studied in the literature before.
 Use this non-semi-simple type Lie algebra $\mathcal{G}\oplus \mathcal{h}\oplus \mathcal{h}^*$, we can find a type of bilinear form as:
$$
\Omega_{ab} = {\begin{pmatrix} 
\Omega_{g_i, g_j} &0 &0 \\
  0 & \Omega_{h_i, h_j} & \mathbb{I}\\
0 & \mathbb{I} & 0
\end{pmatrix}}
$$
each blocks is spanned by generators of $\mathcal{G}$, $\mathcal{h}$, $\mathcal{h}^*$ respectively. But $\Omega_{ab}$ this does not provide us the desired form as $\omega_{ab}$. I suppose we find the Lie algebra by ${f_{ai}}^k\omega_{jk}+{f_{aj}}^k\omega_{ik}=0$ and here $\omega_{ik}=\omega_{ki}$ symmetric for this given $\omega_{ab}$. Say here $[X_i,X_j]={f_{ij}}^k X_k+\dots$. 


*

*Lastly pardon that our understanding is more based on the theoretical physics side. We do have PhDs and do some research works. But our math accents are unfortunately not canonical/orthodox. Please feel free to comment/reply/correct/provide Ref for us. (if you vote down, please provide useful/rational comments on why/how to improve our post.) Thank you.

 A: If I understand your question correctly, then $\mathfrak{g} \ltimes \mathfrak{g}^*_{\textrm{abelian}}$, for $\mathfrak{g}$ any three-dimensional real Lie algebra, has an invariant inner product of this type.  This is the semidirect product of $\mathfrak{g}$ with its coadjoint representation treated as an abelian Lie algebra.
In the generic case, you would have to take the matrix $\Omega = 0$, but for some particular Lie algebras you can allow nontrivial (positive-semidefinite) $\Omega$.  For example, if you take $\mathfrak{g} = \mathfrak{su}(2)$, then you can let $\Omega$ be any (non-positive) multiple of the Killing form.
Since you mentioned the language of double extensions, this is just the double extension of the trivial zero-dimensional Lie algebra by $\mathfrak{g}$.
Added
Let me try to help with the digestion by writing the structure constants explicitly in terms of a basis.  Choose a basis $(X_i)$ for $\mathfrak{g}$ and let $(\alpha^i)$ denote the canonically dual basis for $\mathfrak{g}^*$.  Then the structure constants for $\mathfrak{g} \ltimes \mathfrak{g}^*_{\textrm{abelian}}$ relative to this choice of basis are given by:
$$
[X_i, X_j] = f_{ij}{}^k X_k \qquad\qquad [X_i,\alpha^j] = - f_{ik}{}^j \alpha^k \qquad\qquad [\alpha^i,\alpha^j] = 0
$$
This choice is such that the $a_i$ are all equal.  If you are prescribing the $a_i$ ab initio then you don’t have to choose the canonical dual basis for $\mathfrak{g}^*$: rescaling them should do the trick.
