What is the diameter of the circle inscribed in triangle $ABC$ if $AB = 11,$ $AC=6,$ and $BC=7$? Express your answer in simplest radical form.
Let $d$ be the diameter of the inscribed circle, and let $r$ be the radius of the inscribed circle. Let $s$ be the semiperimeter of the triangle, that is, $s=\frac{AB+AC+BC}{2}=12$. Let $K$ denote the area of $\triangle ABC$.

Heron's formula tells us that \begin{align*}
K &= \sqrt{s(s-AB)(s-AC)(s-BC)} \\
&= \sqrt{12\cdot 1\cdot 6\cdot 5} \\
&= \sqrt{6^2\cdot 10} \\
&= 6\sqrt{10}.
\end{align*}The area of a triangle is equal to its semiperimeter multiplied by the radius of its inscribed circle ($K=rs$), so we have $$6\sqrt{10} = r\cdot 12,$$which yields the radius $r=\frac {\sqrt{10}}{2}$. This yields the diameter $d = \boxed{\sqrt{10}}$.