\mathcal{O}_{S_3}$, and $\mathcal{O}_{S_3,\zeta^2} = \langle x_3^2, x_3y_3, y_3^2 \rangle\mathcal{O}_{S_3}$.

For $P_2 = [0:1:0]$ one uses the change of coordinates on $G_\lambda(x,1,z)$, $(x, z) = (x_2, z_2-x_2)$ to get a polynomial of the form $\beta_\lambda x_2^4 + z_2^3 + \sum_{ij} b_{\lambda ij} x_2^i z_2^j$ with $3i+4j>12$ and $\beta_\lambda \neq 0$. The blow-up with respect to $(3,4)$ resolves the singular point. The exceptional divisor is denoted by $E_2$ and $m_2 = 12$, $\nu_2 = 7$. The second summand in  is $$\label{eq:M_P2}
\frac{\mathcal{O}_{\CC^2}}{\mathcal{M}_{\mathcal{C}_{\lambda},P_2}^{(k)}}
= \frac{\mathcal{O}_{\CC^2}}{ \left\{ g \ \big| \ \operatorname{mult}_{E_3} \pi^{*} g > k - 7 \right\}} \cong 
\begin{cases}
0 & \text{ if }\ \ k = 0,\ldots,6, \\
\CC & \text{ if } \ \ k = 7,8,9, \\
\langle 1,x_2 \rangle_\CC & \text{ if } \ \ k = 10, \\
\langle 1,x_2,z_2 \rangle_\CC & \text{ if } \ \ k = 11.
\end{cases}$$

Finally, for $P_1 = [1:0:0]$ one blows up $G_\lambda(1,y_1, \lambda^2(z_1-y_1^3)) =
\alpha_\lambda y_1^{10} + z_1^3 + \sum_{ij} a_{\lambda ij} y_1^i z_1^j$ with $3i+10j>30$ and $\alpha_\lambda \neq 0$ with weight vector $(3,10)$ to resolve the singular point. The exceptional divisor $E_1$ gives rise to $m_1=30$ and $\nu_1=13$. The first vector space in  is $$\label{eq:M_P1}
\frac{\mathcal{O}_{\CC^2}}{ \left\{ g \ \big| \ \operatorname{mult}_{E_3} \pi^{*} g > \dfrac{5k}{2} - 13 \right\}}
\cong
\begin{cases}
0 & \text{ if } \ \ k = 0,\ldots,5, \\
\CC & \text{ if } \ \ k = 6, \\
\langle 1,y_1 \rangle_\CC & \text{ if } \ \ k = 7, \\
\langle 1,y_1,y_1^2 \rangle_\CC & \text{ if } \ \ k = 8, \\
\langle 1,y_1,y_1^2,y_1^3 \rangle_\CC & \text{ if } \ \ k = 9, \\
\langle 1,y_1,y_1^2,y_1^3,y_1^4,z_1 \rangle_\CC & \text{ if } \ \ k = 10, \\
\langle 1,y_1,y_1^2,y_1^3,y_1^4,z_1, y_1 z_1 \rangle_\CC & \text{ if } \ \ k = 11.
\end{cases}$$ Using the previous calculations one can check that $H^1(Y,\cO_Y(L^{(k)})) = 0$ for any $k \neq 10$ and $\lambda = 1,\zeta$ by studying the cokernel of $\pi^{(k)}$.

To 