Back in the days of the particle zoo, when our baryon count went from proton and neutron to, well, the zoo, people started asking the exact same question: "Why do these particles exist?", and, "Why don't those particles exist?"
The answer is symmetry, and the exact nature of that symmetry led to the quark model and the 1969 Nobel Prize in physics.
The driving factor is the Pauli Exclusion Principle, or, more correctly, the quantum field theory version of it, which is called the Spin-Statistics theorem: fermion states need to change sign if you exchange any pair of particles.
There is little to offer in the realm of intuition as to why that is true. So it's best to use an example.
Consider the deuteron, the simplest nucleus. Here one considers the neutron and proton as identical particles, with a spin-like quantum number (called iso-spin), where the proton is "up", and the neutron is "down". They appear in the deuteron as:
$$ |d\rangle = \frac 1 {\sqrt 2}(|pn\rangle - |np\rangle)$$
Here: $|pn\rangle$ means the 1st particle is a proton and the 2nd particle is a neutron, and likewise for $|np\rangle$.
If you interchange them:
$$ |d\rangle \rightarrow \frac 1 {\sqrt 2}(|np\rangle - |pn\rangle) = -\frac 1 {\sqrt 2}(|pn\rangle - |np\rangle) = -|d\rangle$$
The sign of $ |d\rangle$ changes. That's what antisymmetric under interchange means. (Note also: in deuteron, a single nucleon doesn't have a definite particle identity. It is 50% proton and 50% neutron).
All of these ideas carry over to the up and down quarks of the quark model.
When applied to baryons, it means the total wave function:
$$ \psi = \psi_{\rm space}\psi_{\rm color}\psi_{\rm spin}\psi_{\rm flavor} $$
has to be antisymmetric.
Protons and neutrons are spherical, so $\psi_{\rm space}$ is symmetric (the details follow from the quantum treatment of orbital angular momentum). $\psi_{\rm color}$ is antisymmetric (which is a phenomenological rule of QCD). Thus, the product $\psi_{\rm spin}\psi_{\rm flavor}$ must be symmetric.
How that works for the proton is shown here: Proton spin/flavor wavefunction . (3-particle symmetries in product spaces is involved, but straightforward).
What is germane to your question is this: consider 3 up quarks:
$$ \psi_{\rm flavor} = |uuu\rangle $$
If I exchange any two particles in the flavor state, I get:
$$ \psi_{\rm flavor} \rightarrow |uuu\rangle = +\psi_{\rm flavor} $$
It is symmetric. Hence, the spin wave function must also be symmetric, and when you combined three spin 1/2, the rules of quantum spin tell you the state must be spin 3/2 state, e.g.:
$$ \psi_{\rm spin} = |\uparrow\uparrow\uparrow\rangle $$
So that's it: protons and neutrons are spin 1/2. The symmetry requirements rule out at spin 1/2 stable baryon with 3 up quarks.
You can circumvent this with orbital angular momentum. An $L=1$ state makes $\psi_{\rm space}$ antisymmetric. I think this state is the $N^*(1520)$ listed here: https://pdg.lbl.gov/2019/tables/rpp2019-tab-baryons-N.pdf .
The more well known baryons are the spin-3/2 $\Sigma$ baryons (https://pdg.lbl.gov/2019/tables/rpp2019-tab-baryons-Sigma.pdf), which have a symmetric spin state and a symmetric flavor state.
