Smallest dimension for faithful orthogonal representation $\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$The compact simple Lie groups $\SO_8(\mathbb{R}) $ and $\SO_9(\mathbb{R}) $ both have rank 4. The group
$$
G=\SU_3 \times \SU_2 \times \operatorname{U}_1 
$$
also has rank 4. Does there exist a subgroup of $\SO_8(\mathbb{R}) $ or $\SO_9(\mathbb{R}) $ isomorphic to $ G $?
I already know that there is an inclusion of $ G $ into $\SU_5 $ given by putting $ \SU_3 $ in the top 3 by 3 block, $\SU_2 $ in the bottom 2 by 2 block and then $ U_1 $ corresponds to the block scalar subgroup $\exp(-\frac{2 \pi i t}{3})I_3 \times \exp(\frac{2 \pi i t }{2}) I_2 $ where $ I_n $ are identity matrices of the appropriate size and $ t $ parameterizes the subgroup. Since $ \SU_5 $ naturally embeds in $\SO_{10}(\mathbb{R}) $ we can compose embeddings and get an embedding of $ G $ into $\SO_{10}(\mathbb{R}) $.
 A: I made some mistakes in my first version of this answer (including giving the opposite reply …), so hopefully this is correct.  Thanks to @ZoltanZimboras for help sorting it out (although, of course, if it is still wrong, then the fault is entirely mine).
I claim that there is no embedding of $G$ in $\operatorname{SO}_9(\mathbb R)$ (and hence none in $\operatorname{SO}_8(\mathbb R)$).
If $\{\alpha_1, \alpha_2, \alpha_3, \alpha_4\}$ is a system of simple roots for $\mathsf B_4$, with $\alpha_4$ short, then the subsystems generated by $\{\alpha_1, \alpha_2\}$ and $\{\alpha_4\}$ are orthogonal and of types $\mathsf A_2$ and $\mathsf A_1$, respectively.  I claimed that this gave an embedding of $\operatorname{SU}_3 \times \operatorname{SU}_2$ in $\operatorname{SO}_9(\mathbb R)$, but @ZoltanZimboras pointed out that it actually only gives an embedding in $\operatorname{Spin}_9(\mathbb R)$ (or whatever is the correct notation for the compact spin group).  Note that, under this embedding, the image $\alpha_4^\vee(-1)$ of $-1 \in \operatorname{SU}_2$ in $\operatorname{Spin}_9(\mathbb R)$ lies in the kernel of $\operatorname{Spin}_9(\mathbb R) \to \operatorname{SO}_9(\mathbb R)$, so we cannot simply compose with the projection to get an embedding in $\operatorname{SO}_9(\mathbb R)$.  The additional $\operatorname U_1$ can be taken to be the image of $2\alpha_1^\vee + 4\alpha_2^\vee + 6\alpha_3^\vee + 3\alpha_4^\vee$.
We do not have an embedding of $G$ in $\operatorname{Spin}_8(\mathbb R)$, since the absolute root system of $\operatorname{Spin}_8(\mathbb R)$ is of type $\mathsf D_4$, but $\mathsf D_4$ contains no subsystem of type $\mathsf A_2 + \mathsf A_1$.  (The orthogonal complement of a root in $\mathsf D_4$ is of type $\mathsf A_1 + \mathsf A_1$.)
Suppose we had an embedding $G \to \operatorname{SO}_9(\mathbb R)$.  We claim that some absolute root of $G$ is short in the absolute root system of $\operatorname{SO}_9(\mathbb R)$.  Indeed, suppose not, and consider one of the roots of the $\operatorname{SU}_2$ factor.  Since it is long, its orthogonal complement in $\mathsf B_4$ is of type $\mathsf B_2 + \mathsf A_1$, in Carter's notation (so the $\mathsf A_1$ roots are long).  The long roots in $\mathsf B_2 + \mathsf A_1$ form a system of type $3\mathsf A_1$, which contains no subsystem of type $\mathsf A_2$.
Let $\alpha$ be an absolute root of $G$ that is short in $\operatorname{SO}_9(\mathbb R)$.  Then the simple subgroup of $\operatorname{SO}_9(\mathbb R)$ whose absolute root system is $\{\pm\alpha\}$ is adjoint (i.e., $\operatorname{SU}_2/\langle-1\rangle$), but the simple subgroup of $G$ whose absolute root system is $\{\pm\alpha\}$ is simply connected (i.e., $\operatorname{SU}_2$).  (See @nfdc's answer to Centralizers of subtori in reductive groups, derived subgroups .)  Since these are the same group, we have a contradiction.
@ZoltanZimboras also asked, in a comment on the first version of the answer, how the resulting representations $\operatorname{SU}_3 \to \operatorname{Spin}_9(\mathbb R) \to \operatorname{SO}_9(\mathbb R)$ and $\operatorname{SU}_2 \to \operatorname{Spin}_9(\mathbb R) \to \operatorname{SO}_9(\mathbb R)$ decompose.  I couldn't answer most questions about this representation, but, fortunately, since all computations have been in terms of what happens on tori, I can answer this one!  We have that $(\alpha_1^\vee, \alpha_2^\vee) : (\operatorname S^1)^2 \to \operatorname{Spin}_9(\mathbb R)$ and $\alpha_4^\vee : \operatorname S^1 \to \operatorname{Spin}_9(\mathbb R)$ factor through embeddings of maximal tori $(\operatorname S^1)^2 \to \operatorname{SU}_3$ and $\operatorname S^1 \to \operatorname{SU}_2$, which I will use as coördinates for those tori.
The weights of the (implicit) chosen maximal torus in $\operatorname{Spin}_9(\mathbb R)$, in the natural representation via its projection to $\operatorname{SO}_9(\mathbb R)$, are the trivial weight, together with $\alpha_i + \dotsb + \alpha_4$ and their opposites for $i \in \{1, 2, 3, 4\}$.  Thus the weights of $\operatorname{SU}_3 \times \operatorname{SU}_2 \times \operatorname U_1$ are the trivial weight $(0, 0; 0; 0)$, together with $(1, 0; 0; 2)$, $(-1, 1; 0; 2)$, $(0, -1; 0; 2)$, $(0, 0; 2; 0)$, and their opposites.  Thus, if I haven't mixed up my computation, the restricted representation of $G$ is the sum of the trivial representation; the two fundamental representations of $\operatorname{SU}_3$, on which $\operatorname{SU}_2$ acts trivially, and on one of which the weight of $\operatorname U_1$ is $2$ and on the other of which it is $-2$; and the representation of $\operatorname{SU}_2$ with highest weight $2$, on which $\operatorname{SU}_3 \times \operatorname U_1$ acts trivially.
