Cartan matrix for a semisimple Lie algebra with an extension The question is a modified one inspired by this post:
What is the Cartan matrix for this Lie algebra below? (for this semisimple Lie algebra $g(X) \oplus h(Y)$,)
$$
[X_i, X_j] = f_{ij}{}^k X_k \qquad\qquad [Y_a,Y_b] = F_{ab}{}^c Y_c \qquad\qquad [X_i,Y_a] = \mathcal{F}_{ia}{}^k X_k
$$
Here, $f_{ij}{}^k, F_{ab}{}^c, \mathcal{F}_{ia}{}^k$ are three different structure constants. And $i,j,k \in \{1,2,3\}$, $a,b,c \in \{1,2,3\}$; there are 3 generators $X_1,X_2,X_3$ and 3 generators $Y_1,Y_2,Y_3$.
If there is a generic form of Cartan matrix for this algera will be even better.
If not, we may, for example, consider 3 generators $X_1,X_2,X_3$ generate a compact semi-simple SU(2) Lie algebra with $f_{ij}{}^k$ given by $f_{12}{}^3=1$ and $f_{23}{}^1=-1$ as $i,j,k$ are cyclic. And another 3 generators $Y^1,Y^2,Y^3$ are extension of $X_1,X_2,X_3$. We may also consider also $F_{ab}{}^c$ and $\mathcal{F}_{ia}{}^k $ are structure constants of SU(2) Lie algebra. 
I suppose this modified Lie algebra is still semisimple Lie algebra, because there is no nontrivial maximal solvable Ideal.
Thank you for any comments and concerns. Please provide whatever thoughts!
 A: This depends on the value of ${\mathcal F}$. You will either get direct sum of two su(2)'s or a single su(2). 
Here are some details: I will assume that $f$ and $F$ are structure constants of a simple Lie algebra, say $su(2)$ (using $sl(2)$ will make very little difference). Observe also we have a homomormpism $h: su(2)\to {\mathfrak g}$ with target the Lie algebra ${\mathfrak g}$ you defined, sending generators of $su(2)$ to the vectors $X_i$. Observe that there is no a priori  reason for this homomorphism to be injective; by simplicity of $su(2)$, this homomorphism is either zero or injective. The image $h(su(2))$ is an ideal in ${\mathfrak g}$; dividing by this ideal we obtain Lie algebra with structure constants $F$; by my assumption, this is again $su(2)$. Thus, we obtain a split exact sequence
$$
su(2) \to {\mathfrak g} \to su(2)\to 0. 
$$
This sequence splits (either by assumption as in your question or because $su(2)$ is semisimple); hence, we obtain
$$
 {\mathfrak g}\cong h(su(2))\oplus su(2)
$$
Hence, ${\mathfrak g}$ is either isomorphic to $su(2)$ or to $su(2)\oplus su(2)$. Which case happens depends on ${\mathcal F}$. If indeed ${\mathcal F}$ 
are the structure constants of $su(2)$ then we also obtain a surjection
$$
r: {\mathfrak g}\to su(2)
$$ 
such that $r(X_i)=r(Y_i)$, in which case $h$ cannot be zero. Hence, in this case,
$$
 {\mathfrak g}\cong su(2)\oplus su(2). 
$$
Its Cartain matrix is, of course, just the direct sum of Cartan matrices of two $su(2)$'s. 
