It would be easier to count the number of words that have $2$ or fewer vowels, and then subtract this number from the total number of words (which you have already computed).
I assume that by "$3$" or more vowels" you mean $3$ or more occurrences of vowels, so in particular a word with 2 e's, 2 i's, and the rest consonants qualifies.
How many words are there with no vowels? Clearly
$$21^7$$
if, as per usual convention, we agree that there are $5$ vowels.
How many words with $1$ vowel? Where the vowel occurs can be chosen in $\binom{7}{1}$ ways. For each of these ways, the vowel can be chosen in $5$ ways. And once you have done that, the consonants can be filled in in $21^6$ ways, for a total of
$$\binom{7}{1}(5)(21^6)$$
Finally, how many with $2$ vowels? The location of the vowels can be chosen in $\binom{7}{2}$ ways. Once this has been done, the actual vowels can be put into these places in $5^2$ ways. And then you can fill in the consonants in $21^5$ ways, for a total of
$$\binom{7}{2}(5^2)(21^5)$$
Add up the $3$ numbers we have obtained, subtract from $26^7$.
Our argument was a little indirect. We could instead find, using the same sort of reasoning, the number of words with $3$ vowels, with $4$ vowels, with $5$, with $6$, with $7$, and add up. This is only a little more work than the indirect approach. But any saving of work is helpful! Also, the indirect approach that was described lets us concentrate on pretty simple situations, the most complicated of which is the $2$ vowel case.
Remark: The calculation we have done is closely connected to the Binomial Distribution, and if you have already covered this, it may be the point of the exercise. So if you know about the Binomial Distribution, imagine the letters to be chosen at random. Then the number of patterns with $3$ or more vowels is the probability of $3$ or more vowels, multiplied by $26^7$.
Given an alphabet of $n$ letters one can form $$n\cdot(n-1)\cdots(n-r+1)={n!\over (n-r)!}$$ words of length $r$ using no letter twice.
It follows that there are ${26!\over16!}$ such words of length $10$ from the English alphabet.
There are ${22!\over16!}$ such words of length $6$ not using the letters occurring in ERGO. Each of these words has $7$ spaces (including the ends) where you can insert ERGO.
There are ${21!\over16!}$ such words of length $5$ not using the letters occurring in LATER. Each of these words has $6$ spaces (including the ends) where you can insert LATER.
There are ${19!\over16!}$ such words of length $3$ not using the letters occurring in LATERGO. Each of these words has $4$ spaces (including the ends) where you can insert LATERGO.
Apart from the words containing LATERGO there are no $10$-letter words using no letter twice and containing ERGO as well as LATER.
Using the inclusion-exclusion principle we therefore obtain the following number of words containing no letter twice and not containing ERGO or LATER:
$${26!\over16!}-{22!\over16!}\cdot 7-{21!\over16!}\cdot 6+{19!\over16!}\cdot 4=19\,274\,833\,290\,456\ .$$
Best Answer
Here is simplification. Note that if a five letter word in alphabetical order contains letter $A$, it must start with $A$ and if it contains letter $O$, it must end with $O$. But as we cannot have a vowel at either end, that leaves us to make words with remaining thirteen letters,
B C D E F G H I J K L M NIf the set of words starting with a vowel is $P$ and the set of words ending with a vowel is $Q$,
$|P \cup Q| = |P| + |Q| - |P \cap Q|$
$ \displaystyle |P| = {9 \choose 4} + {5 \choose 4} = 131$
$ \displaystyle |Q| = {7 \choose 4} = 35$
$ \displaystyle |P \cap Q| = 1$
So the answer is $~ \displaystyle {13 \choose 5} - \left(131 + 35 - 1\right) = 1122$