$3\times 3$ 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.
 A: 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 
$$(*)\ \ \ \ \begin{bmatrix} a & b& c \\ \cdot & \cdot & \cdot \\ \cdot& \cdot& \cdot \end{bmatrix},$$
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 & b& c \\ a & \cdot & \cdot \\ \cdot & \cdot & \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
$$\begin{bmatrix} a & b& c \\ a & a& e \\ a& a& f \end{bmatrix},$$
then by the lemma, 
$$0 = C = A = \det \begin{bmatrix} a & e \\ a& f\end{bmatrix}\Rightarrow e = f$$
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 
$$\begin{bmatrix} a & b& c \\ a & b& e \\ a& b& f \end{bmatrix},$$
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 & b& c \\ a & b& c \\ a& b& c \end{bmatrix}
But then 
$$\det \begin{bmatrix} a & a& a \\ b& b& c \\ c & b& c\end{bmatrix} = -a(bc-c^2 ) + a(b^2 - bc)= a(c-b)^2\neq 0$$
This last contradiction concludes the proof.
A: I have given a positive answer for the general case $n\ge3$ in another thread. Below is a scaled-down and simplified proof for the special case where $n=3$. It is produced here to illustrate some ideas of that answer. Let $\mathbf e=(1,1,1)^T$ and
$$
M=\pmatrix{\mathbf x&\mathbf y&\mathbf z}=\pmatrix{a&d&g\\ b&e&h\\ c&f&i}.
$$
Since $M$ has $3$ distinct entries, we may first make $a\ne0$ and make $a,d,g$  distinct. Then at least one of $(a,b,d)^T$ or $(a,b,g)^T$ is not orthogonal to $\mathbf e$. However, as $a\notin\{0,d,g\}$, both $(a,b,d)^T$ or $(a,b,g)^T$ are not parallel to $\mathbf e$. Therefore, by scrambling the entries of $M$ appropriately, we may assume that $\mathbf x$ is neither parallel to nor orthogonal to $\mathbf e$. By permuting the entries of $\mathbf x$ once more, we may further assume that $2a-b-c\ne0$.
For any vector $\mathbf v=(v_1,v_2,v_3)^T$ and any permutation $\sigma\in S_3$, denote $\mathbf v^\sigma=(v_{\sigma(1)}, v_{\sigma(2)}, v_{\sigma(3)})^T$. Since $(2a,\,b+c,\,b+c)^T=(a,b,c)^T+(a,c,b)^T$, all columns of
$$
W=\pmatrix{2a&b+c&b+c\\ b+c&2a&b+c\\ b+c&b+c&2a}
$$
lies inside the linear span of $\{\mathbf x^\sigma:\sigma\in S_3\}$. However, $W$ is nonsingular because its simple eigenvalue $2(a+b+c)=2\mathbf e^T\mathbf x$ and repeated eigenvalue $2a-b-c$ are nonzero. Therefore $\{\mathbf x^\sigma:\sigma\in S_3\}$ spans $\mathbb R^3$.
It follows that $\mathbf y$ and $\mathbf z$ are necessarily linearly dependent. Otherwise, $\{\mathbf u,\mathbf y,\mathbf z\}$ is a basis of $\mathbb R^3$ for some vector $\mathbf u$. However, as $\{\mathbf x^\sigma:\sigma\in S_3\}$ spans $\mathbb R^3$, $\mathbf u=\sum_{\sigma\in S_3}c^\sigma \mathbf x^\sigma$ for some coefficients $c^\sigma$ and we will arrive at a contradiction:
$$
0\ne\det(\mathbf u,\mathbf y,\mathbf z)=\sum_{\sigma\in S_3}c^\sigma\det(\mathbf x^\sigma,\mathbf y,\mathbf z)=0.
$$
Thus we may assume that
$$
M=\pmatrix{\mathbf x&\mathbf y&\mathbf z}=\pmatrix{a&d&td\\ b&e&te\\ c&f&tf}
$$
for some real number $t$. We now scramble the last two columns of $M$ in different ways but keep $\mathbf x$ unchanged. Then $\mathbf y$ and $\mathbf z$ must remain linearly dependent and their $2\times2$ minors must all be zero. In particular
\begin{aligned}
d(e-tf)=\left|\begin{matrix}d&td\\ f&e\end{matrix}\right|=0
\ \text{ and }
\ t(d^2-e^2)=\left|\begin{matrix}d&e\\ te&td\end{matrix}\right|=0.
\end{aligned}
Therefore, by symmetry, we get a system of equations
\begin{align}
&d(e-tf)=d(f-te)=e(f-td)=e(d-tf)=f(e-td)=f(d-te)=0,\tag{1}\\
&t(d^2-e^2)=t(e^2-f^2)=t(d^2-f^2)=0.\tag{2}
\end{align}
The group of equations $(1)$ has four solutions: (a) $d=e=f$ and $t=1$, (b) $d=e=0$, (c) $d=f=0$ and (d) $e=f=0$. So, by $(2)$, we can assume that $M$ is equal to
$$
M_1=\pmatrix{a&d&d\\ b&d&d\\ c&d&d}
\ \text{ or }
\ M_2=\pmatrix{a&d&0\\ b&0&0\\ c&0&0}.
$$
If $M=M_1$, its entries can be rearranged as
$$
\pmatrix{d&a&b\\ d&d&c\\ d&d&d}
$$
and the resulting determinant is $d(d-a)(d-c)$, which by assumption is zero. By cyclically permuting $a,b$ and $c$, we obtain
$$
d(d-a)(d-c)=d(d-b)(d-a)=d(d-c)(d-b)=0,
$$
whose four solutions are $d=0,\,a=b=d,\,a=c=d$ or $b=c=d$. Since $M$ has three distinct entries, the latter three cases are impossible. Therefore $d=0$. 
Thus, regardless of whether $M$ is equal to $M_1$ or $M_2$, it can have at most four nonzero entries. However, since all rearrangements of $M$ in the form of
$$
\pmatrix{a&0&0\\ 0&\ast&0\\ 0&\ast&\ast}
$$
are singular, $M$ can have at most two nonzero entries. Hence the result.
A: If we only aim at solving the special case $n=3$ but not the general case $n\ge3$, there is an easier answer. Let
$$
M=\pmatrix{a&d&g\\ b&e&h\\ c&f&i}
$$
where $a,b,c$ are distinct. From
$$
\det\pmatrix{a&d&g\\ b&e&h\\ c&f&i}=0=\det\pmatrix{c&d&g\\ b&e&h\\ a&f&i}
$$
and Laplace expansion along the first column, we get
$$
a\left|\begin{matrix}e&h\\ f&i\end{matrix}\right|
+c\left|\begin{matrix}d&g\\ e&h\end{matrix}\right|
=c\left|\begin{matrix}e&h\\ f&i\end{matrix}\right|
+a\left|\begin{matrix}d&g\\ e&h\end{matrix}\right|.
$$
Since $a\ne c$, we obtain
$$
\left|\begin{matrix}e&h\\ f&i\end{matrix}\right|
=\left|\begin{matrix}d&g\\ e&h\end{matrix}\right|
\ \Rightarrow
\ \left|\begin{matrix}e&h\\ f-d&i-g\end{matrix}\right|=0,
$$
i.e. $e(i-g)=h(f-d)$. By swapping $f$ and $d$, we also obtain $e(i-g)=h(d-f)$. Therefore $e(i-g)=0$. So, by symmetry, we have $p(q-r)=0$ for any $p,q,r$ choosing from different positions of $(d,e,f,g,h,i)$. It follows that either $d,e,f,g,h,i$ are all equal to each other, or five of them are zero. In other words, we may assume that $M$ is equal to
$$
M_1=\pmatrix{a&d&d\\ b&d&d\\ c&d&d}
\ \text{ or }
\ M_2=\pmatrix{a&d&0\\ b&0&0\\ c&0&0}.
$$
If $M=M_1$, its entries can be rearranged as
$$
\pmatrix{d&a&b\\ d&d&c\\ d&d&d}
$$
and the resulting determinant is $d(d-a)(d-c)$, which by assumption is zero. By cyclically permuting $a,b$ and $c$, we obtain
$$
d(d-a)(d-c)=d(d-b)(d-a)=d(d-c)(d-b)=0,
$$
whose four solutions are $d=0,\,a=b=d,\,a=c=d$ or $b=c=d$. Since $M$ has three distinct entries, the latter three cases are impossible. Therefore $d=0$. 
Thus, when $M=M_1$ or $M_2$, it has at most four nonzero entries. However, since all rearrangements of $M$ in the form of
$$
\pmatrix{a&0&0\\ 0&\ast&0\\ 0&\ast&\ast}
$$
are singular, $M$ can have at most two nonzero entries. Hence the result.
A: 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 $0$.
$$
A = \det
\left(
\begin{array}{ccc}
1 & e^{2\pi \mathbf{i}/3} & e^{4\pi \mathbf{i}/3}\\
e^{2\pi \mathbf{i}/3} & e^{4\pi \mathbf{i}/3} & 1\\
e^{4\pi \mathbf{i}/3} & 1 & e^{2\pi \mathbf{i}/3}
\end{array}
\right) = 0
$$
So we don't need zeros..
