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-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$of its columns). 5 fixed a typo This (and more - namely we would like to know how$A^{-1}$looks like) must be well-known, but we cannot find relevant references. (IMHO I saw it in a commutative algebra/representation theory text...) Update II. If one instead takes$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}$-minors factoring (according to computer) into products of$L_K$as above. More precisely, if a$\binom{n-1}{d}\times\binom{n-1}{d}$-minor$M$misses one$\ell_i$then one arrives to the situation outlined above, which we know how to deal with, thanks to David's answer. Otherwiseit's pretty mysterious: some minors , we still can see that$\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)$, where$J$corresponding to a column of$M$; there are$\binom{n-1}{d+1}$such$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$vanish identically, and others factorthis apparently has to do with the homology of a simplicial complex related to the index sets$J$of its columns). 4 added a more general case Let$\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. (IMHO I saw it in a commutative algebra/representation theory text...) Update. 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$\binom{n-1}{d}\times\binom{n-1}{d}$-minors factoring (according to computer) into products of$L_K$as above. More precisely, if a$\binom{n-1}{d}\times\binom{n-1}{d}$-minor 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 it's pretty mysterious: some minors vanish identically, and others factor. 3 fixed one more typo 2 corrected a typo:$|J|=n-d$, not$d\$