 Solution

Use $t\ge 0$ to label the initial half-line, with $(x,y,z)=(t,-t,1)$ at $s=0$. In the general characteristic from part (i), this sets $a=t$, $b=-t$, $c=1$. The parametric solution surface is thereforeAt $s=0$ this recovers exactly the given data. Moreover $(x_s,y_s,z_s)=(1,1+2s,2)=(1,z,2)$, so the surface follows the required characteristic vector field. The branch defining a classical graph through the data is determined below by restricting to the side of the projection fold containing $s=0$.