x_{0,0} & x_{0,1} & x_{0,2} & x_{0,3} & x_{1,3} \\
x_{0,1} & x_{0,2} & x_{0,3} & x_{1,3} & x_{2,3} \\
x_{0,2} & x_{0,3} & x_{1,3} & x_{2,3} & x_{3,3}
\end{array}\right).$$ The homogeneous saturated ideal of a general $3$-rd hyperplane section of $V$ is generated by the maximal minors of the matrix $$Z=\left(
\begin{array}{cccc}
x_{0,0} & x_{0,1} & x_{0,2} & x_{0,3} \\
x_{0,1} & L_{1,1} & L_{1,2} & x_{1,3} \\
x_{0,2} & L_{1,2} & L_{2,2} & x_{2,3} \\
x_{0,3} & x_{1,3} & x_{2,3} & x_{3,3}
\end{array}\right)$$ where $L_{1,1},L_{1,2},L_{2,2}\in
k[x_{0,0},x_{0,1},x_{0,2},x_{0,3},x_{1,3},x_{2,3},x_{3,3}]$ are general linear forms. Let $I(s)=I_3(Z_s)$ be the ideal generated by the submaximal minors of the matrix $Z_s=sZ+(1-s)Y$. Then one can check that for a generic value of $s$ the cardinality of a minimal system of generators of $I(s)^2$ is $\mu(I(s)^2)=55$ (we used the computer algebra software CoCoA [@cocoa]). If $I(s)$ defines a standard determinantal scheme, then it follows by knowledge of the graded Betti numbers of $I(s)$ that it must be associated to a matrix of linear forms of size $3\times 5$. In that case we have 5 linearly independent Plücker relations, which implies that $\mu(I(s)^2)\leq 50$. Therefore $I(s)$ cannot define a standard determinantal scheme. Hence $V$ has a good determinantal 