When does $\det\left(\begin{smallmatrix}1&1&1\\x_1&x_2&x_3\\x_2&x_3&x_1\end{smallmatrix}\right)=0$ imply $x_1=x_2=x_3$? Original question. Does the condition $$\det\begin{pmatrix}1&1&1\\x_1&x_2&x_3\\x_2&x_3&x_1\end{pmatrix}=0$$ imply $x_1=x_2=x_3$?
a) Is this true if we assume $x_{1,2,3}\in\mathbb R$?
b) Is this true if we assume $x_{1,2,3}\in\mathbb C$?
I believe that this is true for real numbers and false for complex numbers. I posted my arguments in the answer below. (But still I would be interested in any alternative solutions and additional insights.)
We might ask about an arbitrary field - but I would not be surprised it the generalized question might be difficult.
Bonus question. For which fields the above condition holds? If this is difficult to answer, are there are at least some sufficient or necessary conditions?
Some stuff which immediately comes into mind - and perhaps some of these might be reasonable ways to look at this problem.

*

*Let us denote $T(x_1,x_2,x_3)=(x_2,x_3,x_1)$.

*The condition about the determinant basically says that the vectors $\vec u=(1,1,1)$, $\vec x$ and $T(\vec x)$ are linearly dependent.

*Clearly, $T$ is a linear isomorphism and $T^3=id$. If we are working over complex numbers, then the minimal polynomial it $m(t)=t^3-1$ and the eigenvalues are the roots of this polynomial: $1$, $\omega$, $\omega^2$.

*For real numbers, the condition about determinant says that the points $(x_1,x_2)$, $(x_2,x_3)$, $(x_3,x_1)$ are collinear.

*We can forget the linear structure and simply view this as a problem about a polynomial in three variables - we're asking when $x_1x_2+x_2x_3+x_3x_1-x_1^2-x_2^2-x_3^2=0$.

I will add that this question originated from some discussion in chat - although in that discussion it was just an auxiliary result.
 A: We're looking at the condition
$$\det\begin{pmatrix}1&1&1\\x_1&x_2&x_3\\x_2&x_3&x_1\end{pmatrix}=0 \qquad\Rightarrow\qquad x_1=x_2=x_3 \tag{*}$$
a) The condition $(*)$ is true for real numbers.
If this determinant is zero, we have
$$x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_3x_1=\frac{(x_1-x_2)^2+(x_2-x_3)^2+(x_3-x_1)^2}2=0.$$
We can get zero as a sum of non-negative real numbers only if each of them is zero. So we get
\begin{align*}
(x_1-x_2)^2=(x_2-x_3)^2=(x_3-x_1)^2&=0\\
x_1-x_2=x_2-x_3=x_3-x_1&=0
\end{align*}
which means $x_1=x_2=x_3$.
b) The condition $(*)$ is false for complex numbers.
If suffices to check that for $(x_1,x_2,x_3)=(1,\omega,\omega^2)$, where $\omega$ is a third root of unity, we get
$$\det
\begin{pmatrix}
  1 & 1 & 1 \\
  1 & \omega & \omega^2 \\
  \omega & \omega^2 & 1
\end{pmatrix}=0
$$
In fact, it is easy to notice that the rows are linearly dependent, since we get the third row if we multiply the second one by $\omega$.
b') All complex matrices which give zero determinant.
The linear transformation $T(x_1,x_2,x_3)=(x_2,x_3,x_1)$ has three eigenvalues $1$, $\omega$ and $\omega^2$. (The roots of the polynomial $\chi_T(t)=t^3-1$. I.e., we have $\omega^3=1$)

*

*$\vec v_1=(1,1,1)$ is an eigenvector for $\lambda=1$.

*$\vec v_2=(1,\omega,\omega^2)$ is an eigenvector for $\lambda=\omega$.

*$\vec v_3=(1,\omega^2,\omega)$ is an eigenvector for $\lambda=\omega^2$.

Since $\vec v_1$, $\vec v_2$, $\vec v_3$ are eigenvectors corresponding to three different eigenvalues, they are linearly independent and they form a basis. Let us have a look at the vector $\vec x=(x_1,x_2,x_3)$ in this basis.
If $\vec x=c_1\vec v_1+c_2\vec v_2+c_3\vec v_3$, then
$$T(\vec x)=c_1\vec v_1+c_2\omega\vec v_2+c_3\omega^2\vec v_3.$$
So the question whether the vectors $\vec v_1,\vec x,T(\vec x)$ are linearly independent is basically a question about the determinant
$$\det
\begin{pmatrix}
  1 & 0 & 0 \\
  c_1 & c_2 & c_3 \\
  c_1 & c_2\omega & c_3\omega^2 \\
\end{pmatrix}=c_2c_3(\omega^2-\omega)
$$
This determinant is zero if and only if $c_2=0$ or $c_3=0$. So to get a zero determinant, we have to take for $\vec x=(x_1,x_2,x_3)$ an eigenvector for some of the two complex eigenvalues (or the zero vector).
