Solution:

The answer is $\mathbf{( C )}$. Let us first consider a slightly different problem: for now, let us leave the number of houses undetermined, and disregard the assumption that the houses cannot be adjacent. In this case, letting $n$ be the number of houses on the street, we would have $\binom{n}{3}$ ways to choose which houses are inhabited by the three acquaintances and $3!$ ways to assign such houses to them.

At this point, to guarantee that the houses are not adjacent, it suffices to insert one house after the first choice and another after the second choice. Since we know that the houses on the street are 10 in total, this means that the $n$ houses considered before are actually 8 houses. Hence the number we are looking for is
$$\binom{8}{3} \cdot 3! = \frac{6 \cdot 7 \cdot 8}{3!} \cdot 3! = 336$$