Is this eigenstate a tensor product or a direct sum? Is $|\mathbf r\rangle$, the state of being exactly at $\mathbf r$, a direct sum or a tensor product of $|x\rangle$, $|y\rangle$ and $|z\rangle$. The same question for $|\mathbf p\rangle$. Now my attempt is the following:
If it were a direct sum, i.e. $|\mathbf r\rangle = |x\rangle +|y\rangle + |z\rangle$ then the position operator may be $\mathbf{\hat r} = \hat x \oplus \hat y \oplus \hat z$, that is a direct sum of the operators on $X = Y = Z = \mathbb R$, such that
$$
\hat x \oplus \hat y \oplus \hat z:X\oplus Y \oplus Z \to X\oplus Y \oplus Z,
$$
\begin{align}
|x\rangle +|y\rangle + |z\rangle \mapsto &\ (\hat x \oplus \hat y \oplus \hat z)(|x\rangle +|y\rangle + |z\rangle)\\
&= \hat x|x\rangle + \hat  y|y\rangle + \hat  z|z\rangle \\
&= x|x\rangle + y|y\rangle + z|z\rangle
\end{align}
But if it were a tensor product, i.e. $|\mathbf r\rangle = |x\rangle \otimes |y\rangle \otimes |z\rangle$, then $\mathbf{\hat r} = \hat x \otimes \hat y \otimes \hat z$
$$
\hat x \otimes \hat y \otimes \hat z:X\otimes Y \otimes Z \to X\otimes Y \otimes Z,
$$
\begin{align}
|x\rangle \otimes |y\rangle \otimes |z\rangle \mapsto &\ (\hat x \otimes \hat y \otimes \hat z)(|x\rangle \otimes |y\rangle \otimes |z\rangle)\\
&= \hat x|x\rangle \otimes  \hat  y|y\rangle \otimes  \hat  z|z\rangle \\
&= x|x\rangle \otimes  y|y\rangle \otimes z|z\rangle \\
&= xyz\ |x\rangle \otimes  |y\rangle \otimes |z\rangle
\end{align}
which doesn't make a lot of sense in terms of eigenvalues and eigenvectors. So which one is it, or is it none of them ? and should $\mathbf{\hat r} |\mathbf r\rangle = \mathbf r|\mathbf r\rangle$ mean anything ?  See this for more on the definitions of these maps.
 A: It’s a tensor product as the various kets you have live in distinct Hilbert spaces.  In this space $\hat x$ really is $\hat x\otimes \hat{\mathbb{I}}\otimes \hat{\mathbb{I}}$, $\hat y$ is formally $\hat{\mathbb{I}}\otimes \hat y\otimes \hat{\mathbb{I}}$ etc.  Indeed $\vert x\rangle $ is formally $\vert x\rangle \otimes \hat{\mathbb{I}}\otimes \hat{\mathbb{I}}$ and operator like $\hat x\otimes \hat y\otimes \hat z$ acting on $\vert \mathbf{r}\rangle$ would return $xyz\vert \mathbf{r}\rangle$.
A: 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,
$$|\mathbf r\rangle = |x\rangle \otimes |y\rangle \otimes |z\rangle= |x\rangle   |y\rangle  |z\rangle,$$
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 r\rangle$s, as the accepted answer details, $ \hat x  |\mathbf r\rangle =  x|\mathbf r\rangle$, etc. Whence your target 3d vector expression,
$$\mathbf{\hat r} |\mathbf r\rangle = \mathbf r|\mathbf r\rangle,$$
quite meaningful indeed. Yet again, $|\mathbf r\rangle$ 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,
$$   \mathbf{ a}\cdot \mathbf{\hat r} |\mathbf r\rangle = \mathbf {a \cdot r}|\mathbf r\rangle ~;\\   \mathbf{ \hat r}\cdot \mathbf{\hat r} |\mathbf r\rangle =  r^2|\mathbf r\rangle,
$$
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 $|x\rangle$ is a 2-vector;  y to only take 3 positions, so $|y\rangle$ is a 3-vector; and z to only take 4 positions, so $|z\rangle$ is a 4-vector.
Their direct product space $|\mathbf r\rangle$ 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.
