The conic bundle of a cubic threefold Let $X$ be a smooth cubic threefold over $\mathbb{C}$ and let $L \subset X$ be a line.
The projection from $L$ yields a rational map $X \dashrightarrow \mathbb{P}^2$. Resolving the indeterminacy by blowing up $L$, we obtain a conic bundle morphism $\pi: \tilde{X} \to \mathbb{P}^2$.
Consider the relative anticanonical bundle $\omega_{\pi}^{-1}$. By standard properties of conic bundles we obtain an embedding
$$\tilde{X} \hookrightarrow \mathbb{P}(\mathcal{E})$$
which respects $\pi$. Here $\mathcal{E} := \pi_*\omega_{\pi}^{-1}$ is a vector bundle of rank $3$ on $\mathbb{P}^2$ and $\mathbb{P}(\mathcal{E}) \to \mathbb{P}^2$ is the associated $\mathbb{P}^2$-bundle over $\mathbb{P}^2$.

Is there is an explicit description of the vector bundle $\mathcal{E}$? For example, is it independent of $X$ or $L$? Is it determined by the normal bundle of $L$? Is it a direct sum of lines bundles? 

Note that $\mathbb{P}(\mathcal{E}) \cong \mathbb{P}(\mathcal{E} \otimes \mathcal{O}(k))$ for any $k \in \mathbb{Z}$. In particular, I am quite happy with determining $\mathcal{E}$ up to twist by a line bundle.
 A: Choose coordinates $(U,V,X,Y,Z)$ in $\mathbb{P}^4$ so that $L$ is the line $X=Y=Z=0$. The equation of $X$ is of the form
$$ AU^2+2BUV+CV^2+2D U + 2EV +F=0\ ,$$where $A,B,\ldots F$ are homogeneous forms in $X,Y,Z$ of degree $1,1,1,2,2,3$. You can view this equation as a section $s \in H^0(\mathbb{P}^2,\mathrm{Sym}^2\mathcal{E} \otimes \mathcal{O}_{\mathbb{P}^2}(1))$, where $\mathcal{E}$ is the vector bundle $\mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}(1)$. Put $P:=\mathbb{P}_{\mathbb{P}^2}(\mathcal{E})$, and let $p:P\rightarrow \mathbb{P}^2$ be the structure map. Then $s$ defines a section of the line bundle $\mathcal{O}_{P}(2)\otimes p^*\mathcal{O}_{\mathbb{P}^2}(1)$ on $P$, whose zero locus is $\tilde{X} $. 
So up to a twist, your vector bundle is just $\mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}(1)$. You may then compute $\omega _{\pi }$ by the adjunction formula and see which twist is needed to get $\pi _*\omega _{\pi }$.
