Solution in Haskell to puzzle from https://twitter.com/ValdisKrebs/status/819301951972265985

puzzle.hs

import Data.List

let numAndPlace xs ys = length $ (zip xs [0..]) `intersect` (zip ys [0..])
let numNotPlace xs ys = (length $ xs `intersect` ys) - (numAndPlace xs ys)
let test xs ys = (numAndPlace xs ys, numNotPlace xs ys)
let match xs = map (test xs) [[6,8,2],[6,1,4],[2,0,6],[7,3,8],[8,7,0]] == [(1,0),(0,1),(0,2),(0,0),(0,1)]

filter match [[x,y,z] | x<-[0..9], y<-[0..9], z<-[0..9]]

The above can be shortened somewhat if you are prepared to obfuscate! For example:

import Data.List
let l=length;i=intersect;f x=map(\y->(l$i(zip x[0..])(zip y[0..]),l$i x y))["682","614","206","738","870"]==[(1,1),(0,1),(0,2),(0,0),(0,1)];d=['0'..'9']
filter f[[x,y,z]|x<-d,y<-d,z<-d]

You can also shorten the code by removing the " - (numAndPlace xs ys)" and adjusting the desired totals to compensate [(1,1),(0,1),(0,2),(0,0),(0,1)], although I think this makes the configuration less clear. This change has therefore been applied to the obfuscated short version only!