Consider the first number is a then the second is a+1 and the third one is a+2.
according the presentation we have; a(a+1) = 3(a+2) + 2 we solve this equation for a and we get;
a2+a = 3a+6+2 ---->>>> a2-2a-8=0 --->>> (a-4)(a+2) = 0 so a=4 (the negative root -2 is rejected)
so the numbers are 4, 5 , 6
if we take the negatve root it would be -2 , -1 , 0 which gives no results.