$3\sqrt{11}=\sqrt{9}\times \sqrt{11}=\sqrt{9\times 11}=\sqrt{99}$

Similarly $4\sqrt{7}=\sqrt{112}$;

$5\sqrt{5}=\sqrt{125}$;

$6\sqrt{3}=\sqrt{108}$;

$7\sqrt{2}=\sqrt{98}$.

Hence $4\sqrt{7}$, $5\sqrt{5}$ and $6\sqrt{3}$ are all greater than 10.