Is this eigenstate a tensor product or a direct sum?

Just to flesh out @ZeroTheHero 's impeccable answer for you, with a hint of how to escape conceptual hash by small finite-dimensional matrices. I'll avoid addressing your unsound conjectures/probes, to protect your attention from expressions which are not even wrong, in favor of standard stuff.

The states are tensor product states, $$|mathbf rrangle = |xrangle otimes |yrangle otimes |zrangle= |xrangle |yrangle |zrangle,$$ whereas 3d vectors $mathbf r= (x,y,z)^T$ are just that. They can be made into eigenvalues of 3d vectors of operators, $hat {mathbf r}= (hat x,hat y,hat z)^T$, operators acting on the space of $ |mathbf rrangle$s, as the accepted answer details, $ hat x |mathbf rrangle = x|mathbf rrangle$, etc. Whence your target 3d vector expression, $$mathbf{hat r} |mathbf rrangle = mathbf r|mathbf rrangle,$$ quite meaningful indeed. Yet again, $|mathbf rrangle$ is not a vector, much unlike $mathbf{ r}$. It yields the latter under action of the vector $mathbf{hat r} $.

You may further dot this 3-vector equation by a fixed 3-vector $mathbf a,$ to reduce it to just one equation (scalar); or to itself, $$ mathbf{ a}cdot mathbf{hat r} |mathbf rrangle = mathbf {a cdot r}|mathbf rrangle ~; mathbf{ hat r}cdot mathbf{hat r} |mathbf rrangle = r^2|mathbf rrangle, $$ etc.

Your instructor must have taught you how to illustrate such Hilbert spaces by finite-dimensional vector spaces when you are groping for your bearings. Take x to only take 2 positions, so $|xrangle$ is a 2-vector; y to only take 3 positions, so $|yrangle$ is a 3-vector; and z to only take 4 positions, so $|zrangle$ is a 4-vector.

Their direct product space $|mathbf rrangle$ then is 24d, (whereas their direct sum space would be 9d). All operators on this space are thus 24×24 matrices, trivial to visualize. So, if the two eigenvalues of $hat x$ are $x_1$ and $x_2$, do you see the diagonal 24×24 $~~~hat x$ matrix consisting of an upper 12×12 block with entries $x_1$ and a lower 12×12 block with entries $x_2$? Trying to visualize some of your proposed constructions, by contrast, this way, would be simply impossible/ inconceivable--not even wrong. This language should enable you to contrast the 3-vectors of the continuous case, also present and controlling here, to the 24d vector kets.