= Solution
Let $(r_0,r_1,r_2)$ be any ordering of the three distinct numbers $0,1,\lambda$. Set $h=r_1-r_0$ and choose $s\ne0$ with $s^2=h^3$. The <projective transformation>
$$
\begin{pmatrix}X\\Y\\Z\end{pmatrix}
=\begin{pmatrix}
1/h&0&-r_0/h\\
0&1/s&0\\
0&0&1
\end{pmatrix}
\begin{pmatrix}x\\y\\z\end{pmatrix}
$$
has the required matrix form. As $x=hX+r_0Z$ and $y=sY$, substitution gives
$$
Y^2Z=X(X-Z)(X-\widetilde\lambda Z),
\qquad
\boxed{\widetilde\lambda=\frac{r_2-r_0}{r_1-r_0}.}
$$
The choice $s^2=h^3$ is exactly what cancels the coefficient introduced by the change in $x$.
The six orderings give the <Legendre parameter orbit>:
$$
\begin{array}{c|c}
(r_0,r_1,r_2)&\widetilde\lambda\\\hline
(0,1,\lambda)&\lambda\\
(1,0,\lambda)&1-\lambda\\
(0,\lambda,1)&1/\lambda\\
(\lambda,0,1)&(\lambda-1)/\lambda\\
(1,\lambda,0)&1/(1-\lambda)\\
(\lambda,1,0)&\lambda/(\lambda-1)
\end{array}
$$
Every entry is different from $0,1$ when $\lambda\ne0,1$.
There are no further possibilities for a matrix of the stated form. Its <determinant> is $acd$, so $a,c,d$ must be nonzero. On $z=1$ it induces the affine change $u\mapsto(a/d)u+b/d$ and scales $v=y/z$ by $c/d$. It preserves the line $y=0$, whose three intersections with the <plane cubic> have $u$-coordinates $0,1,\lambda$. Their images must have coordinates $0,1,\widetilde\lambda$. Choosing the two roots sent to $0,1$ determines the affine change and leaves exactly the six orderings above.
For general $\lambda$ the six values are distinct. They coincide in some special cases: $\lambda\in\{-1,2,1/2\}$ gives three values, and $\lambda^2-\lambda+1=0$ gives two. In particular, for the value obtained in part (c), the two possible values are $(1\pm\mathrm i\sqrt3)/2$. This does not contradict the paper's qualification that there are six for general $\lambda$.
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