= Solution
At the generic point of $L_0$, the functions $x_1,x_2$ are units and the equation gives $x_0=x_3^3/(x_1x_2)$. Taking $x_3$ as a uniformizer yields
$$
\nu_{L_0}(x_0/x_3)=3-1=2.
$$
At $L_1$, the functions $x_0,x_2$ are units and $x_1=x_3^3/(x_0x_2)$, so $\nu_{L_1}(x_0/x_3)=-1$. Symmetrically, $\nu_{L_2}(x_0/x_3)=-1$.
The complement of the three lines is $U=D_+(x_3)$, with
$$
U\cong\operatorname{Spec}k[u_0,u_1,u_2]/(u_0u_1u_2-1)
\cong\operatorname{Spec}k[u_0^{\pm1},u_1^{\pm1}].
$$
This <unique factorization domain> has trivial divisor class group, so part b says that $[L_0],[L_1],[L_2]$ generate $\operatorname{Cl}(X)$. The units on $U$ are scalar multiples of $u_0^au_1^b$, and their boundary valuation vectors are generated by
$$
(2,-1,-1),\qquad(-1,2,-1).
$$
The resulting integer matrix has <Smith normal form> $\operatorname{diag}(1,3,0)$. Consequently
$$
\operatorname{Cl}(X)\cong
\mathbb Z^3/\langle(2,-1,-1),(-1,2,-1)\rangle
\cong\mathbb Z\oplus\mathbb Z/3\mathbb Z.
$$
Solved by gpt-5.6-sol high.
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