Step-by-Step Solution

1. First, we need to find the values of $n$ for which $3^n - 3$ is a multiple of 7. This means $3^n - 3 \equiv 0 \pmod{7}$, or $3^n \equiv 3 \pmod{7}$.
2. We can simplify this to $3^{n-1} \equiv 1 \pmod{7}$.
3. Now, let's find the powers of 3 modulo 7:
   • $3^1 \equiv 3 \pmod{7}$
   • $3^2 \equiv 2 \pmod{7}$
   • $3^3 \equiv 6 \pmod{7}$
   • $3^4 \equiv 4 \pmod{7}$
   • $3^5 \equiv 5 \pmod{7}$
   • $3^6 \equiv 1 \pmod{7}$
   So, the order of 3 modulo 7 is 6. This means $3^k \equiv 1 \pmod{7}$ if and only if $k$ is a multiple of 6.
4. Therefore, $n-1$ must be a multiple of 6, i.e., $n-1 = 6k$ for some integer $k$. This gives $n = 6k + 1$.
5. We are given that $10 \leq n \leq 100$. Substituting $n = 6k + 1$, we get $10 \leq 6k + 1 \leq 100$.
6. Subtracting 1 from all parts of the inequality, we have $9 \leq 6k \leq 99$.
7. Dividing by 6, we get $\frac{9}{6} \leq k \leq \frac{99}{6}$, which simplifies to $1.5 \leq k \leq 16.5$.
8. Since $k$ must be an integer, the possible values of $k$ are $2, 3, 4, \dots, 16$.
9. The number of integers in this range is $16 - 2 + 1 = 15$.