Let $P_1,\cdots, P_6$ be points in $\mathbb{R}^3.$ Let $D$ be the $6\times 6$ matrix whose $(i,j)$th-entry is the square of the distance between $P_i$ and $P_j$. Show that $\det(D) = 0$.
Note that it suffices to write $D$ as a sum of $5$ matrices of rank at most 1, since for any matrices A and B, $\mathrm{rank}(A+B)\leq \mathrm{rank}(A) + \mathrm{rank}(B) $ and so by induction $\mathrm{rank}(\sum_{i=1}^n A_i) \leq \sum_{i=1}^n\mathrm{rank}(A_i)$. Also $D$ is symmetric since $|P_i - P_j| = |P_j-P_i|$ for all $i,j$.
$D$ can trivially be written as the sum of $6$ matrices with rank at most 1; let a distinct row of D be the first row of each matrix while all other rows are zero. Suppose we have the specific example where $P_1,P_2,\cdots, P_6 = (1,1,1),(1,1,2),(1,2,1),(2,1,1),(2,1,2),(2,2,1).$ Then $D = \begin{pmatrix}0 & 1 & 1 &2 & 2 & 2\\ 1 & 0 & 2 & 2 & 1 & 3\\ 1 & 2 & 0 & 2 & 3 & 1\\ 1 & 2 & 2 & 0 & 1 & 1 \\ 2 & 1 & 3 & 1 & 0 & 2\\ 2 & 3 & 1 & 1 & 2 & 0\end{pmatrix}.$
One might be able to compute $\det(D)$ in this special case by using row or column operations. Also note that the determinant is $n$-linear and alternating.
If $P_i=P_j$ for some $i\neq j$ then $D$ will have two identical rows and hence determinant of zero, so we may suppose that all the $P_i$'s are distinct.
Let $P$ denote the matrix whose columns are $P_1,\dots,P_6$. Let $e,v \in \Bbb R^6$ denote vectors in $\Bbb R^6$ such that $e = (1,\dots,1)$ and $v = (|P_1|^2,\dots,|P_6|^2)$. Show that $$ D = ev^T + ve^T - 2P^TP, $$ which is the sum of 2 rank-1 matrices and one rank-3 matrix.