$3\times 3$ matrix always has determinant $0$. Must $7$ of the elements be $0$? Let $M$ be a $3\times 3$ real matrix with at least $3$ distinct elements and the property that any permutation of it's elements gives a matrix with determinant $0$.
Must $M$ contain exactly seven $0$s?

This question is a special case of my previous question: Exactly $n-1$ nonzero elements if $\det(A)=0$ for every arrangement
Thanks to user Holonomia, we know a counterexample to the analogous question for $2\times 2$ matrices: Any $2\times 2$ matrix with two $1$s and two $-1$s has determinant $0$.
However, I have not been able to make much progress on the $3\times 3$ case. Theoretically the property gives us a system of $9!$ equations (with some symmetries and repeats), but I haven't found a good way to compute with this idea.
Let $a, b, c$ be three distinct elements in the entries of the matrix. We will be using this lemma a lot:

Woolfitt's lemma: Let the matrix be given by 
$$(*)\ \ \ \ \begin{bmatrix} a &amp; b&amp; c \\ \cdot &amp; \cdot &amp; \cdot \\ \cdot&amp; \cdot&amp; \cdot \end{bmatrix},$$

then $A=B=C$, where $A, B, C$ are the minors with respect to $a, b, c$ respectively. 

(See the comment for the proof)

Now we split into two cases. 


First case: there is yet another entry $d$ not equals to $a, b, c$. Then consider the matrix


\begin{bmatrix} d &amp; b&amp; c \\ a &amp; \cdot &amp; \cdot \\ \cdot &amp; \cdot &amp; \cdot \end{bmatrix}

and use the lemma. Then one also get $D' = B' =C'$ (Three new minors). But actually $A = D'$. Thus they are all the same. Then there will be two zeroes in the matrix (the $(3, 2)$ and $(3, 3)$ entries). By my comment in the question, there has to be seven zeroes and we are done. 


Second case: All the other entries are $a, b, c$: By symmetry, assume that we have at least three $a$'s. By another comment below the question, we can assume $a\neq 0$ (or we are done). 


Claim: There are three $b$'s (or $c$): If not, there has to be five $a$'s, but if we expand the first row of

$$\begin{bmatrix} a &amp; b&amp; c \\ a &amp; a&amp; e \\ a&amp; a&amp; f \end{bmatrix},$$

then by the lemma, 
$$0 = C = A = \det \begin{bmatrix} a &amp; e \\ a&amp; f\end{bmatrix}\Rightarrow e = f$$

One can check that this $e$ cannot be $a$ (or there will be seven $a$'s). Then the claim is shown. 

Now we are almost done: the matrix can be written as 

$$\begin{bmatrix} a &amp; b&amp; c \\ a &amp; b&amp; e \\ a&amp; b&amp; f \end{bmatrix},$$

Again using the lemma, we have $e = f$. By the claim, $e=f = c$. (It can't be $a$, $b$). Thus the matrix is 

\begin{bmatrix} a &amp; b&amp; c \\ a &amp; b&amp; c \\ a&amp; b&amp; c \end{bmatrix}

But then 

$$\det \begin{bmatrix} a &amp; a&amp; a \\ b&amp; b&amp; c \\ c &amp; b&amp; c\end{bmatrix} = -a(bc-c^2 ) + a(b^2 - bc)= a(c-b)^2\neq 0$$

This last contradiction concludes the proof.
For $3 \times 3$ we can use three $+1$, three $e^{2\pi \mathbf{i}/3}$ and three $e^{4\pi\mathbf{i}/3}$ and the determinant is $0$.

$$
A = \det
\left(
\begin{array}{ccc}
1 &amp; e^{2\pi \mathbf{i}/3} &amp; e^{4\pi \mathbf{i}/3}\\
e^{2\pi \mathbf{i}/3} &amp; e^{4\pi \mathbf{i}/3} &amp; 1\\
e^{4\pi \mathbf{i}/3} &amp; 1 &amp; e^{2\pi \mathbf{i}/3}
\end{array}
\right) = 0
$$

So we don't need zeros..