We can simply observe that.
Since the product of two rational numbers is always rational, we have that
[tex]0.\bar{7}\cdot \dfrac{1}{3},\quad \dfrac{1}{3}\cdot \dfrac{1}{3},\quad -4\cdot \dfrac{1}{3}[/tex]
are all rationals, since they are the product of two rationals.
On the other hand, we have
[tex]\sqrt{27}=\sqrt{9\cdot 3}=3\sqrt{3}[/tex]
and thus
[tex]3\sqrt{3}\cdot \dfrac{1}{3}=\sqrt{3}[/tex]
which is irrational.
