Certifying $\operatorname{Pic} \overline{X} = \Lambda$

$$\begin{aligned} &Q : a_0 x^2 + a_1 x y + \cdots + a_9 w^2 = 0 \subset \mathbf{P}^3, \quad [L := \mathbf{Q}(\{a_i\}_i):\mathbf{Q}] = 168\\ &\Lambda_Q := \langle [C] : C \subset \sigma(Q) \cap X, \, \sigma : L \hookrightarrow \mathbf{C} \rangle \subseteq \operatorname{Pic}(\overline{X})|_B \subseteq \Lambda \overset{?}{\subseteq} \operatorname{Pic} \overline{X} \end{aligned}$$

The inclusion $\Lambda_Q \subseteq \Lambda$ is not explicit!

Nonetheless, $\operatorname{Pic} \overline{X}$ and $\Lambda$ are saturated in $H_2(X, \mathbf{Z})$.

Hence, it is sufficient to show that $\operatorname{rank} \Lambda_Q = \operatorname{rank} \Lambda = 19$.

We can do this in two ways:

$$\operatorname{Pic} \overline{X} = \Lambda$$