Step 1: Use the sum to fix the middle term.
Three numbers in A.P. can be written as first, middle, last, and their sum is always 3 times the middle term. Since the sum is 27, the middle number is $27/3=9$.
Step 2: Turn the product condition into a quadratic.
Let the first and last numbers be $p$ and $q$. Since the numbers are in A.P., $p+q=2\times9=18$, and we are told $pq=77$. So $p$ and $q$ are the roots of $t^2-18t+77=0$.
Step 3: Solve the quadratic.
$t=\dfrac{18\pm\sqrt{324-308}}{2}=\dfrac{18\pm4}{2}$, giving $t=11$ or $t=7$. So the numbers are $7,9,11$.
\[ \boxed{7,\;9,\;11} \]