Write $a,b,c$ for the sums of each of the three rows. We have $a+b+c=1+\cdots+9=45$. The fact that each row is a multiple of 9 means that $a,b,c\in\{9,18,27,36\}$. So $a,b,c$ are either $27,9,9$ or $18,18,9$, in some order. As $1+2+3+4=10$, the 4-digit term cannot be $9$, so it is $18$ or $27$. But $27$ is not the sum of four consecutive integers, so the third row is $3,4,5,6$ in some order.
We are left with $1,2,7,8,9$ for the other two rows, one adding to $18$ and the other $9$. We cannot reach $18$ with two digits at all or with three digits not using the $9$, so the first row is necessarily one of the following two possibilities: $9,8,1$ in some order, and the middle one $2,7$; or $2,7,9$ in some order and $1,8$. This last case is easily discarded since the only 4-digit product is $279\times18=5022$, which produces the wrong last row.
Looking at the first row, the last digit cannot be $1$ because that would imply a repeated digit. $2$ and $9$ do not fit, because they would put an $8$ in the last row. Actually, if the first row starts with $8$ or $9$, as $819\times27$ already has five digits, we learn that the first number starts with $1$ and that the second one is $27$.
We have $189\times27=5103$, not good, so it has to be $198\times 27=5346$. So the answer to the question is $8+7+6=21$.
Best Answer
If the product is written $abc$ in decimal, then we have$$100a+10b+c=34(a+b+c)$$($a,b,c<10,a\ne0$). Thus$$66a-33c=24b$$and, since $11$ divides the left-hand side, $11|b$. Since $b<10$, $b=0$. We're left with two requirements: $2a=c$ and $34|a0c_{10}$. Testing the possibilities afforded by $2a=c$, we see $102$ is divisible by $34$, so all its multiples are answers also, and the answer is $4$.