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 $8+7+6=21$.
A three-digit positive integer has the form $100h + 10t + u$, where the hundreds digit $h$ satisfies the inequalities $1 \leq h \leq 9$, the tens digit $t$ satisfies the inequalities $0 \leq t \leq 9$, and the units digit
$u$ satisfies the inequalities $0 \leq u \leq 9$. Therefore, we wish to determine the number of solutions of the equation
$$h + t + u = 12 \tag{1}$$
subject to these restrictions.
If we let $h' = h - 1$, then $0 \leq h' \leq 8$. Substituting $h' + 1$ for $h$ in equation 1 yields
\begin{align*}
h' + 1 + t + u & = 12\\
h' + t + u & = 11 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers. A particular solution corresponds to the placement of two addition signs in a row of eleven ones. There are
$$\binom{11 + 2}{2} = \binom{13}{2}$$
such solutions since we must choose which two of the thirteen symbols (eleven ones and two addition signs) will be addition signs.
However, we have counted solutions in which $h' > 8$, $t > 9$, or $u > 9$. We must exclude these.
Suppose $h' > 8$. Then $h'$ is an integer satisfying $h' \geq 9$. Let $h'' = h' - 9$. Then $h'' \geq 0$. Substituting $h'' + 9$ for $h'$ in equation 2 yields
\begin{align*}
h'' + 9 + t + u & = 11\\
h'' + t + u & = 2 \tag{3}
\end{align*}
Equation 3 is an equation in the non-negative integers. Since a particular solution corresponds to the placement of two addition signs in a row of two ones, it has
$$\binom{2 + 2}{2} = \binom{4}{2}$$
solutions.
Suppose $t > 9$. Then $t$ is an integer satisfying $t \geq 10$. Let $t' = t - 10$. Substituting $t' + 10$ for $t$ in equation 2 yields
\begin{align*}
h' + t' + 10 + u & = 11\\
h' + t' + u & = 1 \tag{4}
\end{align*}
Equation 4 is an equation in the non-negative integers with $\binom{3}{2} = 3$ solutions, depending on which variable is equal to $1$. By symmetry, there are also three solutions in which $u' > 9$. No two of these restrictions cannot be violated simultaneously since $9 + 10 = 19 > 12$. Thus, the number of three-digit positive integers with digit sum $12$ is
$$\binom{13}{2} - \binom{4}{2} - 2\binom{3}{2} = 66$$
Best Answer
$\overbrace{111,112,...,119}^{\text{$9$ times}} \quad 141,142,...,149 \quad 191,192,...,199$
$211,212,...,219 \quad 241, 242,...,249\quad291, 292,...,299$
$\quad.\quad\quad\quad\quad\quad\quad\quad\quad.\quad\quad\quad\quad\quad\quad\quad\quad.$
$\quad.\quad\quad\quad\quad\quad\quad\quad\quad.\quad\quad\quad\quad\quad\quad\quad\quad.$
$911,912,...,919 \quad 941,942,...,949\quad991,992,...,999$
So we just need to add up the units, tens and hundreds separately.
We have "$9$ rows" and "$9$ columns" in our first grid (numbers with $1$ as the $2$nd digit).
There are $9\cdot (1+2+3+4+5+6+7+8+9)=9\cdot 45=405$ lots of $100$s. There are also $9\cdot 9=81$ lots of $10$s. We can just focus on the units. There are $9$ rows of $1s$, $9$ rows of $2$s,... and $9$ rows of $9$s. So in total we have $9(1+2+3+4+5+6+7+8+9) = 9\cdot 45 = 405$ from summing all the units. The only difference between each grid is the tens unit.
So in the second grid the tens units sum to $4\cdot 81 \cdot 10 = 3240$.
In the third grid the tens units sum to $9\cdot 81 \cdot 10 = 7290$
Then to find the total we add up all the hundreds:
$40500\cdot 3 =121500$
We add up all the units:
$405\cdot 3 = 1215$
Then we add up all the tens
$810+3240+7290=11340$
The total is then given by $T = 121500+1215+11340=134055$
Thus the answer is $A$