Let $\ell\_1,\dots,\ell\_n$ be $d+1$-variate linear forms over complex numbers in variables $X=(X\_0,\dots,X\_d)$. Consider the $(n-d)$-fold products 
$$\ell\_{i\_1}(X)\ell\_{i\_2}(X)\dots\ell\_{i\_{n-d}}(X)=\sum\_{|I|=n-d}a\_{I,J}X^I,\ J=(i\_1,\dots,i\_{n-d}),\\ 1\leq i\_1<i\_2<\dots < i\_{n-d}\leq n.$$ 


Define $\binom{n}{d}\times\binom{n}{d}$-matrix $A$ with entries $A\_{IJ}=a\_{I,J}$. Then it appears that 
$$ \det A=C\prod\_{K}L\_K,\quad K=(k\_1,\dots,k\_{d+1}),\ 1\leq k\_1< k\_2<\dots<k\_{d+1},$$
where each $L\_K$ is a $(d+1)\times (d+1)$-minor of the $(d+1)\times n$-matrix $L$ of the coefficients of $\ell\_1,\dots,\ell\_n$, and $C\neq 0$.


For instance, let $n=4$, $d=2$, and 
$$L=\left(\begin{array}{rrrr}
a\_{0} & a\_{1} & a\_{2} & a\_{3} \\
b\_{0} & b\_{1} & b\_{2} & b\_{3} \\
c\_{0} & c\_{1} & c\_{2} & c\_{3}
\end{array}\right)$$


Then $$ A = {\scriptsize \left(\begin{array}{rrrrrr}
a\_{0} a\_{1} & a\_{0} a\_{2} & a\_{0} a\_{3} & a\_{1} a\_{2} & a\_{1} a\_{3} & a\_{2} a\_{3} \\
a\_{1} b\_{0} + a\_{0} b\_{1} & a\_{2} b\_{0} + a\_{0} b\_{2} & a\_{3} b\_{0} + a\_{0} b\_{3} & a\_{2} b\_{1} + a\_{1} b\_{2} & a\_{3} b\_{1} + a\_{1} b\_{3} & a\_{3} b\_{2} + a\_{2} b\_{3} \\
a\_{1} c\_{0} + a\_{0} c\_{1} & a\_{2} c\_{0} + a\_{0} c\_{2} & a\_{3} c\_{0} + a\_{0} c\_{3} & a\_{2} c\_{1} + a\_{1} c\_{2} & a\_{3} c\_{1} + a\_{1} c\_{3} & a\_{3} c\_{2} + a\_{2} c\_{3} \\
b\_{0} b\_{1} & b\_{0} b\_{2} & b\_{0} b\_{3} & b\_{1} b\_{2} & b\_{1} b\_{3} & b\_{2} b\_{3} \\
b\_{1} c\_{0} + b\_{0} c\_{1} & b\_{2} c\_{0} + b\_{0} c\_{2} & b\_{3} c\_{0} + b\_{0} c\_{3} & b\_{2} c\_{1} + b\_{1} c\_{2} & b\_{3} c\_{1} + b\_{1} c\_{3} & b\_{3} c\_{2} + b\_{2} c\_{3} \\
c\_{0} c\_{1} & c\_{0} c\_{2} & c\_{0} c\_{3} & c\_{1} c\_{2} & c\_{1} c\_{3} & c\_{2} c\_{3}
\end{array}\right)}$$


and $$ \det A=(-a\_{3}b\_{2}c\_{1}+a\_{2}b\_{3}c\_{1}+a\_{3}b\_{1}c\_{2}-a\_{1}b\_{3}c\_{2}-a\_{2}b\_{1}c\_{3}+ a\_{1}b\_{2}c\_{3}) \\ \times (a\_{3} b\_{2}c\_{0} - a\_{2} b\_{3} c\_{0} - a\_{3} b\_{0} c\_{2} + a\_{0} b\_{3} c\_{2} +
a\_{2} b\_{0} c\_{3} - a\_{0} b\_{2} c\_{3})\\ \times (- a\_{3} b\_{1} c\_{0}+a\_{1} b\_{3} c\_{0} + a\_{3} b\_{0}c\_{1}-a\_{0} b\_{3} c\_{1}-a\_{1}b\_{0} c\_{3} + a\_{0} b\_{1} c\_{3})\\ \times (a\_{2} b\_{1} c\_{0} - a\_{1} b\_{2}c\_{0} - a\_{2} b\_{0} c\_{1} + a\_{0} b\_{2} c\_{1} + a\_{1} b\_{0} c\_{2} - 
a\_{0} b\_{1} c\_{2})\\ 
=L\_{(234)}L\_{(123)}L\_{(124)}L\_{(134)}. $$
This (and more - namely we would like to know how $A^{-1}$ looks like) must be well-known, but we cannot find relevant references. 


**Update II**. If one instead takes $n-d-1$-fold products of $\ell\_i$, then one gets, in the same way, a $\binom{n-1}{d}\times\binom{n}{d+1}$-matrix, with determinants of $\binom{n-1}{d}\times\binom{n-1}{d}$-minors factoring into products of $L\_K$ as above. More precisely, if a $\binom{n-1}{d}\times\binom{n-1}{d}$-minor $M$ misses one $\ell\_i$ then one arrives to the situation outlined above, which we know how to deal with, thanks to David's answer. Otherwise, we still can see that $\det M$ is divisible by $L\_K$, where $K$ is one of $d+1$-subsets of $(1,\dots,n)$ distinct from the complement of $J=(i\_1,\dots,i\_{n-d-1})$ in $(1,\dots,n)$, where is $J$ corresponding to a column of $M$; there are $\binom{n-1}{d+1}=\binom{n}{d+1}-\binom{n-1}{d}$ such $K$. If $L\_K$ vanishes then the forms $\ell\_t$ comprising its columns have a common zero $z$, and as $K\cap J\neq\emptyset$, the vector $(z^I)$ is in the left kernel of $M$. Degree count now shows that $\det M$ factors into the product of $L\_K$.
(Some $\det M$ vanish identically, and this apparently has to do with the homology of a simplicial complex related to the index sets $J$ of its columns).



do you know this determinant (basic commutative algebra)?
Here's a proof; I don't know any references. I will show that, if $L\_K=0$, then $\det A=0$. Since the $L\_K$ are distinct irreducible polynomials, this shows that $\prod L\_K$ divides $\det A$, and the two sides have the same degree.


Suppose that $L\_K=0$. Then the linear forms $\ell\_{k\_1}$, ..., $\ell\_{k\_{d+1}}$ have a common zero; call it $(z\_0, z\_1, \ldots, z\_n)$. Every $I$ has nonempty intersection with $K$, since $|I|+|K|=(n-d)+(d+1) = n+1$. So every product $\ell\_{i\_1} \cdots \ell\_{i\_{n-d}}$ vanishes at $(z\_0, z\_1, \ldots, z\_d)$.


That means that the vector of monomials $(z^J)$ in in the left kernel of $A$, so $\det A=0$.


