When does a cubic surface pass through five lines? The set of 5-tuples of lines in $\mathbf{P}^3$ is parametrized by the 20-dimensional product of Grassmannians $G(2,4)^{\times 5}$.  The set of cubic surfaces is parametrized by a 19-dimensional projective space.
I can present a line in $\mathbf{P}^3$ as a $2 \times 4$ matrix (the projectivization of the row span) and five lines in $\mathbf{P}^3$ as a $10 \times 4$ matrix.  Which rational function of this matrix vanishes on the set of tuples of lines that are contained in a cubic surface?
 A: I would write it differently. First, consider the universal space $M$ consisting of tuples $(S,L_1,\dots,L_5)$, where $S$ is a cubic surface and $L_i$ are lines on $S$. This space lies inside the product $P^{19}\times Gr(2,4)^5$ and can be described as the zero locus of a canonical global section of the vector bundle
$$
E := O(1) \boxtimes (S^3U_1^* \oplus S^3U_2^* \oplus S^3U_3^* \oplus S^3U_4^* \oplus S^3U_5^*)
$$
on it, where $U_i$ is the tautological bundle on the $i$-th copy of the Grassmannian. Consequently, the structure sheaf of $M$ has the following Koszul resolution
$$
0 \to \Lambda^{20}E^* \to \dots \to \Lambda^2E^* \to E^* \to O \to O_M \to 0.
$$
We are interested in the image of $M$ in $Gr(2,4)^5$, so let us push forward the above resolution to $Gr(2,4)^5$. Note that $E^*$ restricts as a sum of $O(-1)$ to any fiber $P^{19}$ of the projection, hence the $p$-th term restricts as a sum of $O(-p)$. Thus restriction of almost all terms are acyclic, and they do not contribute to the pushforward. The only terms which do are the first and the last.
The first gives $O$, and the last gives 
$$
\det(S^3U_1 \oplus \dots \oplus S^3U_5) = \otimes_{i=1}^5\det(S^3U_i) = O(-6,-6,-6,-6,-6).
$$
This means that the equation of the image of $M$ is a hypersurface of polydegree $(6,6,6,6,6)$ on the product of the Grassmannians. Of course its equation is the one given by David Speyer --- the determinant is a homogeneous polynomial of polydegree $(12,12,12,12,12)$ in coefficients of matrices, but if you write it in terms of the Plucker coordinates (which are quadratic in coefficients of matirces), it will have polydegree $(6,6,6,6,6)$.
A: It's a $20 \times 20$ determinant. Take each $2 \times 4$ matrix
$$L = \begin{pmatrix}
s & t & u & v \\
w & x & y & z \\
\end{pmatrix}$$
and turn it into the $4 \times 20$ matrix
$$M := \begin{pmatrix}
s^3 & s^2 t & s^2 u & \cdots & v^3\\
3 s^2 w & 2 swt+s^2x & 2swu+s^2 y & \cdots & 3 v^2 z \\
3 s w^2 & 2 swx+w^2 t & 2swy + w^2 u & \cdots & 3 v z^2 \\
w^3 & w^2 x & w^2 y & \cdots & z^3 \\
\end{pmatrix}$$
If the pattern isn't so clear, the columns are indexed by the $20$ cubic monomials in $4$ variables. For each monomial, plugin $( a \ b ) \begin{pmatrix}
s & t & u & v \\
w & x & y & z \\
\end{pmatrix}$ for the $4$ entries and collect coefficients of $a$ and $b$. For example, the monomial $s^2 t$ becomes $(as+bw)^2 (at+bx)$; expanding and collecting coefficients gives the column $(s^2 t,\ 2swt+s^2 x, \ 2swx + w^2 t, \ w^2 x)^T$. A cubic vanishes on the row span of $L$ if and only if that cubic is in the kernel of $M$. ${}{}{}{}$
Stack up the $5$ $M$ matrices and take the determinant to get your answer.
