Quintic surface singular along lines What is the maximum number of lines, along which a quintic surface in $\mathbb{P}^3$ can be singular ?
 A: The maximum number of lines along which a quintic surface in $\mathbb{P}^3$ can be singular is indeed $6$.
As Francesco Polizzi proved this number is at most $6$. Now, you need to produce a quintic surface that attains this bound.
If $C$ is a plane curve of degree $d$ with $\delta$ nodes then the genus formula yields $\delta \leq \frac{(d-1)(d-2)}{2}$. Furthermore, there is an irreducible plane curve of degree $d$ having $\delta$ nodes as its only singularities if and only if $\delta \leq \frac{(d-1)(d-2)}{2}$, see for instance Theorem 2.1 here https://arxiv.org/pdf/2008.02640.pdf
Now, take $d = 5$ and $\delta = 6$. Then there is an irreducible plane curve $C$ of degree $5$ having exactly $6$ nodes $q_1,\dots,q_6$. Let $S\subset\mathbb{P}^3$ be the cone over $C$ with vertex a point $p\in\mathbb{P}^3$ not lying in the plane spanned by $C$. Then $S$ is an irreducible surface of degree $5$ such that $Sing(S) = L_1\cup\dots\cup L_6$, where $L_i$ is the line spanned by $p$ and $q_i$. Furthermore, $S$ has multiplicity $2$ along $L_i\setminus\{p\}$, and multiplicity $5$ at $p$.
