My attempt: call this set $\Omega$. I feel like we will have to look at $q$ even and $q$ odd separately. Suppose that $q$ is even, then (since the characteristic of the underlying field is 2), we can rewrite the equation for $\Omega$ as $\sum X_i = 0$. So, $\Omega$ is a plane in $\operatorname{PG}(3,q)$. Since the projective space has dimension three, we know that every other plane will intersect $\Omega$ in at least line, proving the statement for even $q$'s.
Now, I don't really know how to handle the case for odd $q$. When we intersect $\Omega$ with a plane $\pi$, we will end up with a quadratic in this plane. Since the field now has odd characteristic, and the matrix of such a quadratic is the identity (indeed, take f.ex. $\pi: X_3 = 0$, then $\Omega\cap \pi: (X_0, X_1,X_2) I_3(X_0,X_1,X_2)^t=X_3=0$), we could consider the corresponding orthogonal polarity* $\beta$ such that $\Omega$ is the set of absolute points of $\beta$. This is where the inspiration ends for me.
$\ast$: a polarity $\beta$ is an anti-automorphism of the projective space for which $\beta^2=1$. An orthogonal polarity in a plane over a field of odd characteristic can be represented as $(a_0,a_1,a_2)^t = A (x_0,x_1,x_2)^t$, where $A$ is symmetric.
Does anyone know how to build on from my reasoning? Or is there another way to tackle this problem?
Thanks!
As you wrote, the intersection is always projectively equivalent to $$x_1^2+x_2^2+x_3^2=0,$$ from here we get that $$x_1^2+x_2^2=-x_3^2.$$ It always has a solution, because in a finite field of odd order every element is the sum of two square-elements. (So for all $-x_3^2$ we can get $x_1$ and $x_2$ this way.)