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 $26^7$.
Hint: One approach is to define four functions. Let $OE(n)$ be the number of words of length $n$ which have an odd number of $1$'s and an even number of $2$'s and three similar ones. Write the recurrence relations between these four. They stay closely balanced, so they are all just about $\frac {3^n}4$ in each one. Maybe you can find a proof of the small correction (note that this is not an integer, which the final result must be)
Added: you are right that $EO(n)=OE(n)$ by symmetry. That is a good thing to think about, as it gets you down to three functions. You need to write a recurrence for each function separately. A word where both are odd can come from a word where both are odd by adding a 3, from EO by adding a 1, or from OE by adding a 2. So $OO(n)=EO(n-1)+OE(n-1)+OO(n-1)$ Your starting condition is $EE(0)=1, \text{others}(0)=0$ because the empty word is even in both. I made a spreadsheet to calculate the first dozen values or so. You can then see a pattern in the corrections, which you can prove from the recurrences. Your final answer is then $EO(n)$
Best Answer
Call a sequence with no two adjacent a's good. A good sequence either (i) ends with b or c or (ii) ends with a.
The Type (i) good sequences of length $n$ are obtained by appending b or c to a good sequence of length $n-1$. So there are $2W(n-1)$ Type (i) good sequences of length $n$.
The Type (ii) good sequences of length $n\ge 2$ are obtained by appending an a to a good sequence of length $n-1$ that doesn't end in a. Such a good sequence is obtained by appending b or c to a good sequence of length $n-2$, so there are $2W(n-2)$ of them.
It follows that for $n\ge 2$, $$W(n)=2W(n-1)+2W(n-2).$$ As initial conditions we can use $W(0)=1$. $W(1)=3$.
Remark: It might be interesting to attempt to use Inclusion/Exclusion to find an expression for $W(n)$. Doable, but we end up with a messy sum.
The recurrence we found is by no means the only possible one. However, it is a linear homogeneous recurrence with constant coefficients, so it can be solved using standard tools.