Take $\mathbb{P}^2$ and blow up three points on the same line, let the resulting surfaces be $S$. Use $H$ to denote the hyperplane and $E_i,i=1,2,3$ to denote the exceptional divisors. Consider the complete linear system $|3H-E_1-E_2-E_3|$, gives a morphism $f:S\to \mathbb{P}^6$ where the image $f(S)$ is a degree $6$ singular del Pezzo obtained by contracting the $-2$ curve $H-E_1-E_2-E_3$ on $S$.
My question: if we consider the projection of $\mathbb{P}^6$ from the double point on $f(S)$, what would be a minimal compactification of the image of $f(S)$ in $\mathbb{P}^5$?
Clearly the three exceptional curves $E_i$ will be contracted by the projection since they are straight lines through the double point.
Thanks!
First you have to blowup the singular point of the sextic del Pezzo; then what you get is your surface $S$; the exceptional divisor of this blowup is the unique curve $E$ in the linear system $H - E_1 - E_2 - E_3$. Then the projection is given by the linear system $$ (3H - E_1 - E_2 - E_3) - E = (3H - E_1 - E_2 - E_3) - (H - E_1 - E_2 - E_3) = 2H. $$ The result, of course, is just the Veronese surface in $\mathbb{P}^5$.