Inscribed circle in a right-angled triangle

Let $h$ be the altitude of the triangle, whose line splits $\sqrt{170}$ into $x$ and $\sqrt{170}-x$. We know that the total area of the triangle is $$\frac12\cdot7\cdot11=\frac12h(\sqrt{170}-x)+\frac12hx\implies\text{altitude}=h=\frac{77}{\sqrt{170}}.$$

I assume you mean that $CH$ is the altitude from $C$ to the hypotenuse.

Since $\triangle ABC$ is a right triangle, and since $\triangle AHC$ and $\triangle HBC$ each share one angle with $\triangle ABC$, the three triangles are similar. We have $$ \frac{CH}{AC} = \frac{BC}{AB} $$ and (equivalently) $$ \frac{CH}{BC} = \frac{AC}{AB}. $$

The radii of the three incircles are proportional to $AB,$ $AC,$ and $BC,$ so their sum is \begin{align} \frac{a+b-c}{2} + \frac{a(a+b-c)}{2c} + \frac{b(a+b-c)}{2c} &= \frac{(a+b+c)(a+b-c)}{2c} \\ &= \frac{(a+b)^2 - c^2}{2c} \\ &= \frac{ab}{c}, \end{align} which is indeed equal to the altitude.