Because the probability of each die coming up $5$ is different, we can't short-cut this by using combinatorics. We have to consider all the possibilities.
Let's call the probability that a die comes up $5$ "$F$", and not-$5$ "$G$". Then, let's label each of the dice with a suffix of $1$, $2$, or $3$.
Of course, the $G$ value for each die is $1$ minus the $F$ value.
As you said "exactly $2$", then we must consider all the combinations with exactly $2$ $F$'s and $1$ $G$. Namely:
$$P = F1 \times F2 \times G3 + F1 \times G2 \times F3 + G1 \times F2 \times F3$$
$$= 0.7 \times 0.48 \times (1-0.38) + 0.7 \times (1-0.48) \times .38 + (1-0.7) \times 0.48 \times 0.38$$