smc

Let the edge lengths be $a,b$, and $c$. Therefore, the total area of all its faces is $2ab+2bc+2ac$. Therefore, $2ab+2bc+2ac=22$ and $ab+bc+ac=11$. Also, the total lengths of all of its edges is $4a+4b+4c$, so $4a+4b+4c=24$, and $a+b+c=6$. Therefore, we have: $ab+bc+ac=11$ and $a+b+c=6$. The length of the interior diagonal is $\sqrt{a^2+b^2+c^2}$, so if we can find $a^2+b^2+c^2$, we can find the diagonal. We square the second equation above to introduce an $a^2+b^2+c^2$: $(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ac=36$. However, we already know that $2ab+2bc+2ac=22$, so $a^2+b^2+c^2+22=36⟹a^2+b^2+c^2=14⟹\sqrt{a^2+b^2+c^2}=\sqrt{14},\boxed{\sqrt{14}}$.