First let us draw the generalized Venn Diagram of the situation:
As per the question the portion whose cardinality we need to find is as shown below in blue:
The required cardinality can be given as
$$N=|V\cap F| +|V\cap E|+ |F \cap E| - 2|V\cap E\cap F|$$
But we know :
$$|V\cup E\cup F|=|V|+|E|+|F|-|V\cap F| -|V\cap E|- |F \cap E| +|V\cap E\cap F|$$ $$ = 9+7+10 -|V\cap F| -|V\cap E|- |F \cap E| +|V\cap E\cap F|$$ $$ = 26 -|V\cap E\cap F| -N = 21- N \text{ [since $|V\cap E\cap F|=5$]}$$
$$\implies N=21-|V\cup E\cup F|$$
The minimum possible value of $|V\cup E\cup F|$ is $10$ as we could possibly pack the sets $V$ and $E$ in $F$ as shown below (since $F$ is having the largest cardinality among all the three sets): [This is possible because, no explicit values of $|V\cap F|$,$|V\cap E|$ or $|F \cup E|$ are given, so we are free to think of any possible sort of overlapping]
So the least value of $|V\cup E\cup F|$ is $10$.
So the max value of $N$ is $11=(21-|V\cup E\cup F|=21-10)$
But the minimum value of $N$ is $5$ since it is given that $|V\cap E\cap F|=5$, and the diagram shall be as shown:
So $$5\leq N \leq 11$$