Let $a_0,a_1,a_2$ be rationals, $\det\begin{pmatrix}a_0&a_1&a_2\\a_2&a_0+a_1&a_1+a_2\\a_1&a_2&a_0+a_1\end{pmatrix}=0$, show $a_0=a_1=a_2=0$ 
Let $a_0,a_1,a_2$ be rationals, $\det\begin{pmatrix}a_0&a_1&a_2\\a_2&a_0+a_1&a_1+a_2\\a_1&a_2&a_0+a_1\end{pmatrix}=0$, show $a_0=a_1=a_2=0$.

This is a problem in PHD candidate to PKU (2019).
The matrix is not invariant under permutations of $1,2,3$. So I could not just let the last column being the linear combination of the first two. Soundly, I have no idea. Any hint is grateful.

HINT:
I suggest first to permute $a_1$ and $a_2$, getting the matrix
$$A=\begin{pmatrix}a_0&a_2&a_1\\a_1&a_0+a_2&a_1+a_2\\a_2&a_1&a_0+a_2\end{pmatrix}$$
Now, consider the ring
$$\mathbb{Q}[x]/(x^3-x-1)$$
Since the polynomial $x^3 - x -1$ is irreducible, we get in fact a field.
The multiplication by an element $a_0 + a_1 x + a_2 x^2$ has the above matrix in the basis $1$, $x$, $x^2$.
$\bf{Added:}$ Another solution in the style of linear algebra.
Consider the matrix
$$X=
\begin{pmatrix}0 & 0 & 1\\
1 & 0& 1\\
0 & 1 & 0\end{pmatrix}$$
( the companion matrix of the polynomial $x^3 - x -1$)
Then our matrix equals
$$A = a_0 + a_1 X + a_2 X^2$$
Let's diagonalize $X$. Consider $\alpha$, $\beta$, $\gamma$ the roots of the polynomial $x^3 - x -1$. Then one can check that
$$ \begin{pmatrix}1 & \alpha & \alpha^2\\
1 & \beta & \beta^2\\
1  & \gamma  & \gamma^2 \end{pmatrix} \cdot X = \operatorname{Diag}(\alpha, \beta, \gamma) \cdot \begin{pmatrix}1 & \alpha & \alpha^2\\
1 & \beta & \beta^2\\
1  & \gamma  & \gamma^2 \end{pmatrix}$$
Therefore, $X$ is diagonalizable with eigenvalues $\alpha$, $\beta$, $\gamma$, and $A$ is diagonalizable with eigenvalues $a_0 + a_1 \alpha + a_2 \alpha^2, \ldots$. We conclude that
$$\det A= (a_0 + a_1 \alpha + a_2 \alpha^2)(a_0 + a_1 \beta + a_2 \beta^2) ( a_0 + a_1 \gamma + a_2 \gamma^2)$$
Now we use that $x^3 - x -1$ is irreducible over $\mathbb{Q}$, so neither of $\alpha$, $\beta$, $\gamma$ is root of a rational polynomial of degree $\le 2$.
Let's note that the determinant equals the norm of the element $a_0 + a_1 x + a_2 x^2$ in the field $K = \mathbb{Q}[x]/(x^3 - x -1)$
$$N^K_{\mathbb{Q}}(a_0 + a_1 x + a_2 x^2) = \det A $$
$\bf{Added:}$  Another solution:
Note that we may assume $a_0$, $a_1$, $a_2$ integers (with $\gcd = 1$). We used the polynomial $x^3 - x -1$ which is irreducible over $\mathbb{Q}$. But the polynomial is also irreducible $\mod 2$. We could have worked over the basic field $\mathbb{Z}/2$ instead of $\mathbb{Q}$ and still get the same result. This suggests that we can show the result $\mod 2$. Therefore, we can show: if $a_0$, $a_1$, $a_2$ are integers, not all even then the above determinant is odd. Indeed the determinant equals
$$\det A = a_0^3 + 2 a_0 ^3 a_3 - a_0 a_1^2 - 3 a_0 a_1 a_2 + a_0 a_2^2 + a_1^3 - a_1 a_2^2 + a_3^3 $$
(see this WA calculation). Now, if the $a_i$ are in $\mathbb{Z}/2$ then the expression equals
$$a_0 + a_1 + a_2 + a_0 a_1 + a_0 a_2 + a_1 a_2 + a_0 a_1 a_2 = (1+a_0)(1+ a_1)(1+a_2) + 1$$
and this equals $1 \pmod 2$ if at least one of the $a_i$ is $1 \pmod 2$.  We are done.
