I think I solved it, but my way is far from quick or easy.
for the uncool solution:
Let p1 and p2 be the two primes and s be the perfect square.
p1 + p2 + s = 2009.
since both p1 and p2 are odd, s must be odd aswell.
so s has a range from 10^2<s<32^2 (cannot be higher than 1000)
or to be more precise: s = 11^2 (121), 13^2(169), 15^2(225), 17^2(289), 19^2(361), 21^2(441), 23^2(529), 25^2(625), 27^2(729), 29^2(841) or 31^2(961)
but since the digits cannot be the same:
s = 13^2(169), 17^2(289), 19^2(361), 23^2(529), 25^2(625), 27^2(729), 29^2(841) or 31^2(961)
The only way I can see how to find the numbers with a bit of ease, is to set s really low and then force the primes to be high and therefore I don't have to search through many primes, seeing as there are less primes the higher you get.
the lowest perfect square with the conditions is 169, so set s = 169 and put it in the equation
p1 + p2 + 169 = 2009
p1 + p2 = 1840
let p1>p2
then p1>1840/2(920)
The closest prime to 920 with the conditions is 937, but if p1=937 then p2 is no prime.
we can see
920<p1<1000
and then:
800<p2<900
so substitute p1 = q1 + 900 and p2 = q2 + 800
then
q1 + 900 + q2 + 800 = 1840
q1+q2 = 140
try for q1>q2
then q1 > 140/2(=70)
and the only two primes with a q1 being more than 70 is 971 and 983.
Luckily, 1980 - 983 = 857, which also is a prime.
the primes are: 857, 983
the perfect square is: 169
But I think I know another way. I'ma see if it's easier and then post it later.