Embedding of a blow-up In $\mathbb{P}^1\times\mathbb{P}^2$ take a general divisor $X$ of type $(0,2)$. Consider two general divisors $H_1,H_2$ of type $(2,1)$ and set $Y = X\cap H_1\cap H_2$.
Let $Z$ be the blow-up of $X$ along $Y$. I would like to ask whether there is a rank $3$ vector bundle $\mathbb{E}$ on $\mathbb{P}^1$ such that $Z$ can be embedded as a divisor in $\mathbb{P}(\mathbb{E})$ and if so how the splitting type of $\mathbb{E}$ can be computed.
Thank you very much.
 A: The map $Z \to \mathbb{P}^1$ is a conic bundle, so to understand the vector bundle $\mathbb{E}$ it is enough to compute the pushforward of the anticanonical class.
Now, the anticanonical class of $Z$ can be written as $-K_X - E$, where $E$ is the exceptional divisor of the blowup, so its pushforward to $X$ is isomorphic to $I_Y(-K_X)$. If you use the identification of $X$ with $\mathbb{P}^1 \times \mathbb{P}^1$, this is $I_Y(2,2)$. Using the Koszul resolution for $I_Y$, one obtains
$$
0 \to \mathcal{O}(-2,-2) \to \mathcal{O}^{\oplus 2} \to I_Y(2,2) \to 0.
$$
Pushing this forward to the first factor, one obtains
$$
0 \to \mathcal{O}^{\oplus 2} \to p_{1*}(I_Y(2,2)) \to \mathcal{O}(-2) \to 0.
$$
Therefore, $\mathbb{E}$ has splitting type $(0,0,-2)$ or $(0,0,2)$, depending on which convention about the projective bundle you use.
A: This isn’t exactly what you asked for but it’s close enough that maybe it is good enough.
First of all, $X \cong \mathbb{P}^1 \times \mathbb{P}^1$ and under this isomorphism (where the second factor was a plane conic), the other two divisors become divisors “of type $(2,2)$” on $X$ itself.
Lemma. If $X$ is a variety and $Z \subset X$ is a complete intersection of codimension $d$, then $Z$ embeds as a divisor in $\mathbb{P}(E)$ for some rank $d$ vector bundle $E$ on $X$.
To see this, let $L_1, \ldots, L_k$ be the line bundles and $s_1, \ldots, s_k$ the sections cutting out $Z$. We can think of this as a single section $s$ of $E = \bigoplus_{i=1}^k L_i$, vanishing along $Z$.
This gives a rational section $X \dashrightarrow \mathbb{P}(E)$ by $x \mapsto (x, \langle s(x)\rangle)$ and the closure of the image is $\mathrm{Bl}_Z X$.
So for your blowup, since both equations for $Z$ come from the same line bundle, $E = \mathcal{O}_{\mathbb{P}^1 \times \mathbb{P}^1}(2,2)^{\oplus2}$. Since $\mathbb{P}(E \otimes L) \cong \mathbb{P}(E)$ for all vector bundles $E$ and line bundles $L$, we can twist down to make $E$ trivial. So this embeds your blowup in $X \times \mathbb{P}^1 = (\mathbb{P}^1)^3$.
Not exactly what you asked for but perhaps it’ll be useful.
