Inspired By Brian Charlesworth!

$a^{2} + b^{2} = c^{2}$ $(a+b)^{2} -c^{2} = 2ab$ $(a+b+c)(a+b-c) = 20(a+b+c)$ $a+b-c = 20$ $10(a+b+c) + 10(a+b-c) = ab +200$ $(a-20)(b-20)=200$

Solving the above diophantinus equation and substituting the values of $a$ and $b$ in $a^{2} + b^{2} = c^{2}$ . We get,

$\small S = \left\{ (21,220,221),(22,120,122),(24,70,74),(25,60,65),(28,45,53),(30,40,50)\right\}$

So, $x = 221+122+74+65+53+50 =\boxed{585}$ , $y=\boxed{6}$ .

Now, inradius, $r = \dfrac{\Delta}{s} = \dfrac{\dfrac{1}{2}ab}{\dfrac{1}{2}(a+b+c)} = \dfrac{ab}{a+b+c} = 10$ . So, it is constant for all the triangles formed by these Pythagorean triplets. So, $z=\boxed{10}$ .

So, $x+y+z=\boxed{601}$ .

For a right-angled triangle with sides $a,b,c$ where $a<b<c$ , By Pythagoras theorem we have hypotenuse $c=\sqrt{ a^2+b^2}$ . So our equation becomes $ab = 10\left(a + b + \sqrt{a^2 + b^2} \right)$ $ab -10a - 10b = 10\sqrt{a^2 + b^2}$ . Now, squaring both the sides, we get: $a^{2}b^{2}+ 100(a^{2}+b^{2})-20ab(a+b)+200ab=100(a^{2}+b^{2})$ => $ab(ab - 20a - 20b + 200) = 0$ . Since $ab\neq 0$ , $(ab - 20a - 20b + 200) = 0$ . Factorizing and rearranging the given equation, we get: $a(b-20)-20(b-10)=0$ => $\dfrac{a}{20}=\dfrac{b-10}{b-20}=> a= 20+\dfrac{200}{b-20}$ We can easily solve this Diophantine equation, starting with the factorization of $200$ , followed by setting values for $b$ , then finding $a$ and finally $c$ , under the following constraints: $0<a<b<c$ and $(a,b,c)\in Z^{+}$ . So we arrive at 6 possible solutions of $(a,b,c)$ : $(30,40,50), (28,45,53), (25,60,65),(24,70,74),(22,120,122),(21,220,221)$

Therefore, $x = \sum{c_{\Delta}}=585 , y=6$ . For finding the in-radius, we can apply the formula, $r=\dfrac{a+b-c}{2}$ (This can be derived easily through the properties of tangents to circles). By replacing values from the solution-set we get the in-radii of all triangles as $10$ . $\therefore z=10$ . Hence, $x+y+z= 585+6+10=601 :)$

$\text{Corollary:}$ Given the equation $ab= 10(a+b+c)$

$=> 10= \dfrac{ab}{a+b+c} => \dfrac{\dfrac{(a+b)^{2}}{2}-\dfrac{(a^{2}+b^{2})}{2}}{a+b+c}$

$=> 10 = \dfrac{(a+b)^{2}-c^{2}}{2(a+b+c)}=> 10= \dfrac{(a+b+c)(a+b-c)}{2(a+b+c)}$

$\therefore 10 = \dfrac{a+b-c}{2}=r$

Hence for any triplet of $(a,b,c)$ satisfying the above equation, the inradius of all possible triangles formed will always be equal to $10$