Let $Z_1,Z_2$ and $Y$ be subvarieties of a locally complete intersection variety $X$ over $\mathbb C$.
Consider the strict transforms of $Z_1$ and $Z_2$ in the blowup $Bl_YX$, the question is: when will their intersection be the strict transform of $Z_1 \cap Z_2$?
The answer in general is no. Even if $Y$ is a Cartier divisor, we don't always have $\overline{Z_1-Y} \cap \overline{Z_2-Y}= \overline{Z_1 \cap Z_2-Y}$ in $X$. For example, $X=\mathbb A^2$ with $Y=\{y=0\}$, $Z_1=\{x=0\}, Z_2=\{x=y\}$.
Another counterexample is $\mathbb A^2$ blowing up at the origin $(x,y)$, and $Z_1, Z_2$ are two curves intersecting tangently at the origin.
Note the universal property of blow up doesn't show the natural map $ Bl_{Z_1\cap Z_2\cap Y}(Z_1\cap Z_2) \rightarrow Bl_{Z_1\cap Y}Z_1\cap Bl_{Z_2\cap Y}Z_2$ is an isomorphism, because the testing objects are limited. It's an isomorphism iff restriction of the exceptional divisor to $Bl_{Z_1\cap Y}Z_1\cap Bl_{Z_2\cap Y}Z_2$ is still a Cartier divisor (recall empty is also a Cartier divisor).
In practice, one can check the question by hand. I am interested in putting some conditions on $Z_1, Z_2, Y$ and $X$ to make the answer true.
Let me assume: all $Z_1,Z_2,Y,X$ are equi-dimensional, and $\dim(Z_1 \cap Z_2)=\dim X - \dim Z_1 - \dim Z_2$; $Z_1 \cap Z_2 \not \subseteq Y$ and $Y \not \subseteq Z_1 \cap Z_2$; $Z_1$ is smooth and irreducible; $Bl_Z X$ is smoooth...
The question is (not directly) related to: if $f \in A$ s.t $\bar{f} \in A/I_1, A/I_2$ are both non-torsion, then is $f \in A/I_1+I_2$ non-torsion? This is also false in general.
