Fix an algebraically closed field $k$ of characteristic 0. Consider an $n$-tuple $(A_1,\ldots, A_n)$ of 
 $n\times n$ matrices over $k$ and assign to it the determinantal surface in $\mathbb{P}_k^{n-1}$ cut out by the polynomial $\det(x_1A_1+\ldots x_nA_n).$ Now consider a fixed non-zero matrix $G$ of rank $r<n$.

Now let $x$ be the vector with components $x_i$ and $G_{ij}$ be the components of $G$. We may then consider the variety cut out by the polynomial $\det(\sum\limits_{j=1}^nx_jA_iG_{ij}).$ With $y=Gx$, this is then a plane section of the variety cut out by $\det(y_1A_1+\ldots y_nA_n)$ with the $\mathbb{P}_k^{r-1}$ corresponding to the image of $G$, and also a determinantal hypersurface of degree $n$ itself in said $\mathbb{P}_k^{r-1}$. 

Now fix a 'generic' $n$-tuple $(A_1,\ldots,A_n)$. 'Generic' means I don't exactly know yet in which sense, but let's say that at least all the matrices are invertible. My question is then whether a generic determinantal hypersurface of degree $n$ in $\mathbb{P}_k^{r-1}$ is a linear section of the determinantal hypersurface of degree $n$ in $\mathbb{P}_k^{n-1}$ corresponding to $(A_1,\ldots,A_n)$.  Actually, even knowing whether some fixed determinantal hypersurface has infinitely many nonisomorphic plane sections would be huge for me. This does not seem like a strong statement, but it's non-obvious for me.