Find the number of ways of arranging the letters

Ignore the C's for the moment and arrange the remaining letters. There are $\frac{12!}{5!3!2!}=332640$ ways to do this.

Now consider the thirteen spaces between and beyond the letters – at most one C may be inserted into each space, so given an arrangement of the other letters there are $\binom{13}3=286$ ways to add C's.

The total number of admissible combinations is thus $332640×286=95135040$.

Try a simpler approach, start with counting he number of ways you can arrange all letters except C's. it's:

$$ \frac{12!}{5!3!2!}$$

now there are 13 places to put the C's where they won't be next to each other so the answer is:

$$\frac{12!}{5!3!2!} \binom{13}{3}$$

While I prefer the approach explained by Alaleh A and Parcly Taxel, here is how you can solve the problem using the Inclusion-Exclusion Principle.

The number of distinguishable arrangements of $5$ A's, $3$ B's, $3$ C's, $1$ D, $2$ E's, and $1$ F is $$\frac{15!}{5!3!3!1!2!1!}$$ as you found.

From these, we must subtract those arrangements in which a pair of C's are adjacent.

A pair of C's are adjacent: We have $14$ objects to arrange, $5$ A's, $3$ B's, $1$ CC, $1$ C, $1$ D, $2$ E's, and $1$ F. They can be arranged in $$\frac{14!}{5!3!1!1!1!2!1!}$$ ways.

However, if we subtract $\frac{14!}{5!3!1!1!1!2!1!}$ from the total, we will have subtracted too much since we will have subtracted each arrangement with two pairs of adjacent C's twice (such arrangements have three consecutive C's), once when we designated the first two C's as the adjacent pair and once when we designated the last two C's as the adjacent pair. Since we only want to subtract such arrangements once, we must add them back.

Two pairs of adjacent C's: As mentioned above, this means the three C's are consecutive. Hence, we have $13$ objects to arrange, $5$ A's, $3$ B's, $1$ CCC, $1$ D, $2$ E's, and $1$ F. The number of such arrangements is $$\frac{13!}{5!3!1!1!2!1!}$$ as you found.

Hence, the number of arrangements of $5$ A's, $3$ B's, $3$ C's, $1$ D, $2$ E's, and $1$ F in which no two C's are consecutive is $$\frac{15!}{5!3!3!1!2!1!} - \frac{14!}{5!3!1!1!2!1!} + \frac{13!}{5!3!1!1!2!1!}$$