I have tried to generalize the Exercise referenced by Karl, even though he told me that it shouldn't be possible this way. I think, however, it works:
Edit: I made a mistake concerning $J_i$ - it cannot be equal to $I_i\oplus\bigoplus_{d\ge 1} I_i^dT^d$ because that is not necessarily an ideal - it might not be closed under multiplication by elements from the ring $S$. The version below looks better.
Proposition. Let $Z_0,\ldots,Z_r$ be closed subschemes of a Noetherian scheme $X$ such that $Z_i\not\subset Z_j$ for $i\ne j$. Let $I_i:=I(Z_i)$ and denote by $\tilde{Z}_i$ the respective strict transform of $Z_i$ under the blow-up $\beta:\tilde{X}\to X$ of $X$ along $I:=\sum_{i=0}^rI_i$. Then, $\bigcap_{i=0}^r\tilde{Z}_i=\emptyset$.
Proof. The statement can be checked locally, so we may assume that $X=\mathrm{Spec}(A)$ is affine. Let $f_i:Z_i\hookrightarrow X$ be the respective closed immersion, so $Z_i=\mathrm{Spec}(A/I_i)$ and $f_i^\sharp:A\twoheadrightarrow A/I_i$. Then, the inverse image ideal sheaf of $I$ under $f_i$ is $I\cdot A/I_i$ and hence,
$\displaystyle\tilde{Z}_i=\mathrm{Proj}\left(\bigoplus_{d\ge 0} \left(I\cdot A/I_i\right)^d\cdot T^d\right)$With $S=\bigoplus_{d\ge 0} I^d\cdot T^d$, the homogeneous ideal defining $\tilde{Z}_i$ inside $\tilde{X}=\mathrm{Proj}(S)$ is equal to
$\displaystyle J_i = \bigoplus_{d\ge 0} (I^d\cap I\cap I_i)$In particular, $J_0+\cdots+J_r\supseteq S_+$, so any point $P\in\tilde{Z}_0\cap\cdots\cap\tilde{Z}_r$ would correspond to a homogeneous prime ideal containing each of the $J_i$ and hence, the irrelevant ideal. There is no such point.
Did I miss something? Or is this correct?
