Homomorphic encryption for vector addition
A very simple way to do something like this is by using the following construction:

$$E^*(\langle x,y,z \rangle) = \langle E(x), E(y), E(z) \rangle,$$

where $E$ is an additively homomorphic cryptosystem.  (To avoid attacks, use three different public keys to encrypt the three components.)  Now $E^*$ has your desired homomorphic property.

I don't understand what the hyperlinear business is, so I've ignored that part.

The problem with this construction is that the three components are separable.  For instance, given $E^*(\langle x,y,z \rangle)$, it is possible to recover $E^*(\langle x,0,0 \rangle)$.  Or, given $E^*(\langle x_1,y_1,z_1 \rangle)$ and $E^*(\langle x_2,y_2,z_2 \rangle)$, it is possible to compute (say) $E^*(\langle x_1+x_2,y_1,z_1 \rangle)$.  These are values that it might not be possible to derive with just $\oplus$.

So, can we do better?  That's what I'll answer next.

Can we do better?

The answer is probably, it depends.  There are a bunch of situations where it is simply not possible to do better -- it's not possible to achieve what you want.  To identify whether it will be possible to achieve what you want, we need to dig a bit deeper into the specific details of exactly what you want, in greater precision.

Suppose we want an encryption scheme

$$E^*: \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/q\mathbb{Z} \times \mathbb{Z}/r\mathbb{Z} \to R$$

that is additively homomorphic on $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/q\mathbb{Z} \times \mathbb{Z}/r\mathbb{Z}$.  In other words,

$$E^*(\langle x_1,y_1,z_1 \rangle)\oplus E^*(\langle x_2,y_2,z_2 \rangle) = E^*(\langle x_1+x_2,y_1+y_2,z_1+z_2 \rangle),$$

and $\oplus$ is the only method we have for combining ciphertexts.

If $p,q,r$ are three different numbers that are pairwise relatively prime, it is definitely not possible to do what you want.  Here's why.  Given $E^*(\langle x,y,z \rangle)$ and an integer $n$, it is always possible to efficiently compute $E^*(\langle nx,ny,nz \rangle)$ using just $\oplus$ (use repeated doubling).  Set $n_1=r \times (r^{-1} \bmod q)$.  Then given $E^*(\langle x,y,z \rangle)$, we can compute $E^*(\langle n_1 x,y,0 \rangle)$ via the formula

$$E^*(\langle n_1 x,n_1 y,n_1 z \rangle) = E^*(\langle n_1 x,y,0 \rangle),$$

using that we know how to get the left-hand side of the formula above from $E^*(\langle x,y,z \rangle)$ and $n_1$.  Next, set $n_2 = q \times (q^{-1} \bmod p) \times (n_1 \bmod p)$.  Now given $E^*(\langle n_1 x,y,0 \rangle)$, we can compute $E^*(\langle  x,0,0 \rangle)$ using the formula

$$E^*(\langle n_2 n_1 x,n_2 y,0 \rangle) = E^*(\langle x,0,0 \rangle).$$

So in this case, given $E^*(\langle x,y,z \rangle)$, we were able to compute $E^*(\langle x,0,0 \rangle)$ using just the operation $\oplus$.  This shows that if $p,q,r$ are mutually co-prime, then it's guaranteed that the three components will be separable, no matter what we do.  So, in this case you might as well use the very simple construction shown at the top of my answer.

What if $p=q=r$?  This looks like one case where it might be possible to achieve your desired property.  However, we have to be a bit careful.  If we know ciphertexts $E^*(\langle x_1,y_2,z_3 \rangle),\dots,E^*(\langle x_k,y_k,z_k \rangle)$ where the linear span of these three vectors is all of $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$, then we can separate the three components and we're back to something that is no better than my very simple construction at the top of the answer.

In conclusion, it's only going to be possible to achieve what you want if you pick $p,q,r$ very carefully and if you severely limit what ciphertexts are available to the eavesdropper.  Moreover, the specific construction might depend on your specific choice of $p,q,r$.  Did you have a strategy to deal with these challenges?  You might want to edit your question to tell us about what values of $p,q,r$ you had in mind and how they'll be chosen and how they're related (mutually different, mutually co-prime, all the same, whatever).