 Solution

The complement $Y$ is covered by the $m$ principal affine opens $D(f_1),\dots ,D(f_m)$. Every finite intersection is again a principal affine open, so this is an acyclic cover for ${\mathcal{O}}_Y$. Its Čech cochain complex has no degree-$m$ term because there is no intersection of $m+1$ distinct cover members. Therefore

Translate $p$ to the origin. The complement ${\mathbb{A}}_k^3\setminus \{0\}$ has the affine cover $D(x),D(y),D(z)$, and its second Čech cohomology isThe class of $x^{-1}{y}^{-1}{z}^{-1}$ is nonzero, so $H^2({\mathbb{A}}_k^3\setminus \{p\},\mathcal{O})\ne 0$. On the other hand, ${\mathbb{A}}_k^3\setminus \ell =D(x)\cup D(y)$, and the first part with $m=2$ gives $H^2({\mathbb{A}}_k^3\setminus \ell ,\mathcal{O})=0$. Since sheaf cohomology is invariant under scheme isomorphism, the two complements are not isomorphic.