Let $p\_1,...,p\_n\in\mathbb{P}^3$ be general points, and let $\Delta\subset\mathbb{P}^3$ be a general surface of degree $d$ with points of multiplicity $m\_i$ at $p\_i$ for $i = 1,...,n$.


Consider the blow-up $X$ of $\mathbb{P}^{3}$ at $p\_1,...,p\_n$ and the strict transform $\tilde{\Delta}$ of $\Delta$. For which $d,m\_1,...,m\_n,$ is $(X,\tilde{\Delta})$ a klt pair?


For instance $m\_1 = ... = m\_2 = 0,1$ will work, and I guess $m\_1 = ... = m\_2 = 2$ as well.



Kawamata-Log-Terminal pairs
I'm not sure what you mean by the pair being klt here: if $\tilde{\Delta}$ appears with coefficient $1$, then it's not. On the other hand, it's sensible to ask whether you can find a boundary $\mathbb Q$-divisor $D$ numerically equivalent to $\tilde{\Delta}$ for which $(X,D)$ is klt. Or you could ask for $(X,\tilde{\Delta})$ to be dlt/lc/plt, as Karl Schwede suggests. In any case, 
I doubt you can really say much in general -- it's not even possible to say for which $d$ and $m\_i$ the class is effective (cf. the Nagata conjecture on $\mathbb P^2$). I guess if the strict transform is basepoint free you're in business, but this probably isn't easy to read off from the numbers.


I'm not sure about your examples: even with $d=1$ and $m\_1=m\_2=1$ I don't think it's klt -- no matter how you write $\Delta$, it's a sum $\sum a\_i P\_i$, with $P\_i$ a plane through the points and $\sum a\_i = 1$, and the divisor you get when you blow up the line has discrepancy $-1$.


