# Lines on Fermat cubic in projective space

I started to learn algebraic geometry by myself not long ago. This question arised from the notes by Andreas Gathmann on Algebraic geometry. I do not understand why we are working in projective space.
In chapter 11, the author introduced the 27 lines on Fermat cubic. But a Fermat cubic is of the form $$V(x^3+y^3+z^3-1)$$ in $$\mathbb{C}^3$$. In the notes, the author did not explain the situation in $$\mathbb{C}^3$$ but in $$\mathbb{C}P^3$$ instead.
Note that in $$\mathbb{C}P^3$$ a line corresponds to a 2-dimensional subspace of $$\mathbb{C}^4$$.
So I think there should be a bijection between the set of lines on $$V(x^3+y^3+z^3-1)$$ and the set of lines on $$V_+(x^3+y^3+z^3+w^3)$$. I tried to prove the bijectivity in the following way
Given a line $$l$$ in $$V_+(x^3+y^3+z^3+w^3)$$, according to the notes, it can be written in the form $$\begin{bmatrix} 1 & 0 & 0 & -\omega^j\\ 0 & 1 & -\omega^k & 0 \\ \end{bmatrix}$$ for $$\omega$$ a primitive 3rd root of unity and $$j,k=0,1,2$$. We take $$j=k=1$$ and now the line has this expression $$(s:t:-\omega t:-\omega s), (s:t)\in\mathbb{C}P^1$$ If we let the last coordinate be $$-1$$, then $$-\omega s=-1$$ means $$s=\omega^2$$. The line is $$(\omega^2:t:-\omega t:-1)$$ It satisfy $$(\omega^2) ^3+t^3+(-\omega^3 t^3)+(-1)=0$$, which is $$1+t^3-t^3-1=0$$. If we consider the line $$(\omega^2,t,-\omega t)\subseteq\mathbb{C}^3$$ we do have a line on $$V(x^3+y^3+z^3-1)$$.
On the other hand, if we have a line $$l$$ on $$V(x^3+y^3+z^3-1)$$, I do not know how to get a line in $$\mathbb{C}P^3$$. But I think since we want a 2-dimensional subspace in $$\mathbb{C}^4$$, so when we think of the set $$\{(x_l,y_l,z_l,w)|w\in\mathbb{C},(x_l,y_l,z_l)\in l\}$$ we should have a plane but it is not a 2-dimensional subspace because it does not pass origin. Then how do we prove the bijection?
Thank you really for helping and reading.