Here, we are drawing all $9$ pens. So,
Number of different ways $ =$ Number of permutations of $9$ objects where $3$ are of type $A$, next $3$ of type $B$ and final $3$ of type C
$$=\frac{9!}{3!3!3!}$$
Now, when we have two red pens in succession, this is equivalent to $2$ objects being close to each other in permutation. So, ignoring these two we can arrange the rest $7$ in $\frac{7!}{3!3!}$ ways. Now, this group of $2$ can be placed in $8$ positions around the $7$ objects -- wait -- here is the problem. A $\text{RED}$ pen is already there and if we put $\text{RED,RED}$ on its left or right side we get the same arrangement. So, effectively we have only $7$ unique positions for the $2$ $\text{RED}$ pens. This gives our required number of permutations as $\frac{7!.7}{3!3!}.$
Thus our required probability will be
$$ \frac{7!.7}{3!3!} \div \frac{9!}{3!3!3!} = \frac{3!.7}{8.9} = \frac{7}{12}$$