Step 1: Another route
Instead of expanding along row 1, we find the determinant by row reduction and then use the signed minor for the required entry.
Step 2: Determinant by row operations
Add 3 times row 1 to row 2 and subtract 2 times row 1 from row 3. This gives rows $(1,3,-2)$, $(0,9,-11)$, $(0,-1,4)$. So $|A|=1\cdot(9\cdot4-(-11)(-1))=36-11=25$.
Step 3: The required entry
The (1,2) entry of the inverse equals the signed minor of position (2,1) divided by $|A|$. Minor of (2,1): delete row 2, column 1 to get $\begin{vmatrix}3&-2\\5&0\end{vmatrix}=0+10=10$. The sign of position (2,1) is negative, so the cofactor is $-10$.
Step 4: Divide
$(A^{-1})_{12}=\frac{-10}{25}=-\frac{2}{5}$, option (A).
Final Answer:
The required entry is $-\frac{2}{5}$.
\[ \boxed{-\dfrac{2}{5}} \]