INTERMEDIATE : Not talking about Level. Part-III

Ways to arrange all vowels = $\frac{6!}{2!3!}=60$ Ways to arrange all consonants = $\frac{6!}{2!}=360$

Since all vowels must be placed together, we have 7 ways to do that, i.e starting from 1st, 2nd,...7th place.

Words can be made = $60\times360\times7=151200$

There are a total of $6$ vowels which can be considered as a single unit. Then the total number of possibilities for these vowels to exist consecutively would be $7!/2!$ as the letter $T$ is repeated twice.
However, the vowels are not the same and there would be various ways to arrange them. The number of possibilities for this would be $6!/(3!*2!)$ since $E$ is repeated thrice and $I$ twice.
Hence we get the to total number of possibilities on multiplying the above possibilities , that is $\boxed{151200}$ .

It Can Be Simply Done Using The So Called Tie Method. We Can Group All The Vowels Together i.e.(I,I,E,E,E,A) And treat them as a single unit.then we can permute this unit with the remaining 6 letter OUT OF WHICH THERE ARE TWO T's So we there need to use concept of permutation of alike objects.and finally we need to multiply the result my internal permutations of the vowel unit where we again need to use permutation of alike objects. on solving the factorial boils to 151200