Vowels are $A,E,A$
Consonants are $L,G,B,R$
$XLXGXBXRX$
Since there is no restriction on consonants, we must place them first and they can be arranged in $4!=24$ days.
Now to arrange vowels so that no two of them are together we have $5$ places $\left(X\right).$ Out of these $5$ places, $3$ can be selected in ${\mathrm{}}^5{C}_3$ ways and then one arrangement can be made in $\frac{3!}{2!}$ ways.
Thus, required ways $=4!\times {\mathrm{}}^5{C}_3\times \frac{3!}{2!}=720$