Question 708125
Starting with three even integers, make the first one 2n, where n in any integer.  n may be odd or even.  Starting with 2n, this is made as an even number. 
The next two numbers will be 2n+2 and 2n+4.


Arranging for the other descriptions:
2n+(2n+4) = the sum of the first one and the third one;
2(2n+(2n+4))= TWICE the sum of the first one and the third one;

21+(2n+2) = Twenty one more than the second integer.


Putting everything together according to the phrasing:
{{{2(2n+(2n+4))=21+(2n+2)}}}
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I have worked through this TWICE now, and have obtained n=5/2.  This means that the first "even" integer is 2(5/2)=5.  This is NOT even.  
I believe something is wrong with the problem description.