The operation of computing the remainder of a division of $a$ by $b$ is called modulo, and written $\text{mod}$. An important property of the modulo operation is that it commutes with addition and multiplication, in a way. You have that $ \begin{eqnarray} (a+b) \text{ mod } c &=& ((a \text{ mod } c) + (b \text{ mod } c)) \text{ mod } c \\ (a*b) \text{ mod } c &=& ((a \text{ mod } c) * (b \text{ mod } c)) \text{ mod } c \end{eqnarray} $
Restated in these terms, the term you want to evaluate is $ 3\cdot16! + 2 \text{ mod } 17$ With the rules above you can simplify that to $ (3\cdot(16! \text{ mod } 17) + 2) \text{ mod } 17 $ Now, to evaluate $16! \text{ mod } 17$, you can either work recursively, i.e. split the product in half, evaluate each side (by using the same splitting technique) and then multiplying the result and applying module again. Or you can use the theorem that $(p-1)! \text{ mod } p = p-1$ if $p$ is a prime. With the second method, you immediate get $ 3\cdot 16 + 2 \text{ mod } 17 = 16 $