Convert three parametric planes into normal and Cartesian forms

With directions $\mathbf u=(1,2,1)$ and $\mathbf v=(1,3,4)$,

$\begin{aligned}n_x&=2\cdot4-1\cdot3=5,\\n_y&=1\cdot1-1\cdot4=-3,\\n_z&=1\cdot3-2\cdot1=1.\end{aligned}$

Thus $\mathbf u\times\mathbf v=(5,-3,1)$.

Use the base point $(1,2,3)$ to obtain $d=(1,2,3)\cdot(5,-3,1)=5-6+3=2$. Expanding the dot product gives the Cartesian form. This plane is the one from Example 28(i).

The source uses $(0,0,2)$ as the first direction here. Its cross product with $(1,-1,2)$ is

$\begin{aligned}n_x&=0\cdot2-2(-1)=2,\\n_y&=2\cdot1-0\cdot2=2,\\n_z&=0(-1)-0\cdot1=0.\end{aligned}$

Thus the cross product is $(2,2,0)=2(1,1,0)$.

Choose the simplified normal $(1,1,0)$. The base point gives $d=(1,2,4)\cdot(1,1,0)=3$, so $x+y=3$. There is no $z$ term: all real $z$ values occur on this vertical plane. The scale change from Example 28(ii)’s direction $(0,0,1)$ does not change the plane.

The two directions have cross product

$\begin{aligned}n_x&=0\cdot3-0\cdot0=0,\\n_y&=0\cdot0-(-1)3=3,\\n_z&=(-1)0-0\cdot0=0.\end{aligned}$

Thus the cross product is $(0,3,0)=3(0,1,0)$.

Choose normal $(0,1,0)$. Then $d=(3,1,2)\cdot(0,1,0)=1$. Both supplied directions have zero $y$ component, confirming that every generated point has $y=1$. The second direction is $(0,0,3)$ in this printed question; it is different from Example 28(iii)’s second direction and defines a different plane.