If we have a configuration of the numbers that has the required property, then we can add or subtract the same number from each of the numbers in the circles and maintain the property. (This is because there are the same number of circles in each line.) Therefore, we can subtract 2012 from all of the numbers and try to complete the diagram using the integers from 0 to 8. We label the circles as shown in the diagram, and call $S$ the sum of the three integers along any one of the lines. Since $p, q, r, t, u, w, x, y, z$ are 0 through 8 in some order, then $p+q+r+t+u+w+x+y+z=0+1+2+3+4+5+6+7+8=36$. From the desired property, we want $S=p+q+r=r+t+u=u+w+x=x+y+z$. Therefore, $(p+q+r)+(r+t+u)+(u+w+x)+(x+y+z)=4S$. From this, $(p+q+r+t+u+w+x+y+z)+r+u+x=4S$ or $r+u+x=4S-36=4(S-9)$. We note that the right side is an integer that is divisible by 4. Also, we want $S$ to be as small as possible so we want the sum $r+u+x$ to be as small as possible. Since $r+u+x$ is a positive integer that is divisible by 4, then the smallest that it can be is $r+u+x=4$. If $r+u+x=4$, then $r, u$ and $x$ must be 0,1 and 3 in some order since each of $r, u$ and $x$ is a different integer between 0 and 8. In this case, $4=4S-36$ and so $S=10$. Since $S=10$, then we cannot have $r$ and $u$ or $u$ and $x$ equal to 0 and 1 in some order, or else the third number in the line would have to be 9, which is not possible. This tells us that $u$ must be 3, and $r$ and $x$ are 0 and 1 in some order. Therefore, the value of $u$ in the original configuration is $3+2012=2015$.