Step 1: List the possible three-digit factorisations of 70.
We need three whole numbers between 0 and 9 whose product is $70$. Start by picking the hundreds digit and see what is left for the other two digits.
Step 2: Try each possible first digit.
If a digit is $1$, the other two digits must multiply to $70$. Checking pairs from $1$ to $9$: no pair of single digits multiplies to $70$, since the closest we can get is $9 \times 9 = 81$ and $70$ has no factor pair both under $10$ (its factor pairs are $1\times70$, $2\times35$, $5\times14$, $7\times10$, none with both members under $10$). So no digit set contains a $1$.
If a digit is $2$, the other two must multiply to $70/2 = 35 = 5 \times 7$. Both $5$ and $7$ are valid digits, so $\{2,5,7\}$ works.
If a digit is $5$, the other two must multiply to $70/5 = 14 = 2 \times 7$, again giving the same set $\{2,5,7\}$.
If a digit is $7$, the other two must multiply to $70/7 = 10 = 2 \times 5$, once more giving $\{2,5,7\}$.
No other starting digit (3,4,6,8,9) divides $70$ exactly with an integer digit-pair result, so this search confirms only one digit set exists.
Step 3: Add the digits.
$2 + 5 + 7 = 14$, and this sum does not change no matter which of the three digits sits in the hundreds place.
Step 4: Match with the options.
Out of $12, 14, 16, 18$, only $14$ equals this sum.
\[
\boxed{14}
\]