Let $$A=\pmatrix{x&a&b\\a&y&c\\b&c&z}$$ be a symmetric matrix (over $\mathbb{R}$ if necessary) such that $\det A=0$ and that $$\left|\matrix{x&a\\a&y}\right|+\left|\matrix{x&b\\b&z}\right|+\left|\matrix{y&c\\c&z}\right|=0$$ (those $2\times2$ matrices are obtained from $A$ by erasing the diagonal elements). Prove that $\operatorname{rank}A\leq 1$.
My attempts: (1) There are $6$ variables and $2$ equations, leaving $4$ degrees of freedom, which is already a bad thing. (2) I wrote down the Laplace extension into the complementary minors, and tried to prove that all the complementary minors are equal to $0$ (which would imply that desired result), but at the end I got $6$ minors versus $4$ equations: $$m_{11}+m_{22}+m_{33}=0$$ $$xm_{11}-ym_{22}-zm_{33}+2cm_{23}=0$$ $$xm_{11}-ym_{22}+zm_{33}+2bm_{13}=0$$ $$-xm_{11}-ym_{22}+zm_{33}+2am_{23}=0$$ where $m_{ij}$ denotes the determinant of $A$ substracting the $i$th row and the $j$th column.
Since $A$ has zero as eigenvalue, its characteristic polynomial is $$\lambda^3-\mathrm{tr}(A)\lambda^2 +\alpha \lambda ,$$ where $\mathrm{tr}(A)=x+y+z$ and $\alpha$ is to be determined. We need to show that $\alpha =0$. But actually, if you do the computation of $\det (\lambda I -A)$ you will see $$-\alpha= \left|\matrix{x&a\\a&y}\right|+\left|\matrix{x&b\\b&z}\right|+\left|\matrix{y&c\\c&z}\right|=0.$$ And we're done.