Solution:

The required integers are 102, 204, 306, 408.
Indeed, let us denote by $a, b, c$, respectively, the hundreds digit, the tens digit, the units digit of a 3-digit integer. The given condition then translates into $100 a+10 b+c=34(a+b+c)$, from which, after simple algebraic steps, we obtain $11(2 a-c)=8 b$.
From this it follows that $b$ must be a multiple of 11, but since $0 \leq b \leq 9$, the only possibility is that $b=0$, and consequently also $2 a=c$. Substituting for $c$ the values $1,2,3,4$ one obtains the 4 integers listed above, which one easily verifies are indeed solutions of the problem. For $c=0$ or $c \geq 5$, on the other hand, no 3-digit integers are obtained.