26. 阅读材料：若$m^{2}-2mn+2n^{2}-8n+16= 0$,求m,n的值.
解：$\because m^{2}-2mn+2n^{2}-8n+16= 0$,$\therefore (m^{2}-2mn+n^{2})+(n^{2}-8n+16)= 0$,$\therefore (m-n)^{2}+(n-4)^{2}= 0$,$\therefore (m-n)^{2}= 0$,$(n-4)^{2}= 0$,$\therefore n= 4$,$m= 4$.
根据你的观察,探究下面的问题：
(1)已知$a^{2}+6ab+10b^{2}+2b+1= 0$,则$a= $,$b= $；
(2)已知等腰三角形ABC的三边长a,b,c都是正整数,且满足$2a^{2}+b^{2}-4a-6b+11= 0$,求$\triangle ABC$的周长；

(3)已知$x+y= 2$,$xy-z^{2}-4z= 5$,求xyz的值.

(1)
$\because a^{2}+6ab+10b^{2}+2b+1 = 0$
$\therefore a^{2}+6ab + 9b^{2}+b^{2}+2b + 1 = 0$
$\therefore (a + 3b)^{2}+(b + 1)^{2}= 0$
$\therefore a + 3b = 0$，$b + 1 = 0$
解得$b=-1$，$a = 3$
(2)
$\because 2a^{2}+b^{2}-4a-6b + 11 = 0$
$\therefore 2a^{2}-4a + 2+b^{2}-6b + 9 = 0$
$\therefore 2(a - 1)^{2}+(b - 3)^{2}= 0$
$\therefore a - 1 = 0$，$b - 3 = 0$
解得$a = 1$，$b = 3$
①当$a$为腰长时，三边为$1$，$1$，$3$，$1 + 1\lt 3$，不能构成三角形，舍去；
②当$b$为腰长时，三边为$3$，$3$，$1$，$3+3\gt 1$，$3 + 1\gt 3$，能构成三角形，周长为$3 + 3+1 = 7$
(3)
$\because x + y = 2$，则$y = 2 - x$
$\because xy-z^{2}-4z = 5$
$\therefore x(2 - x)-z^{2}-4z = 5$
$\therefore 2x-x^{2}-z^{2}-4z - 5 = 0$
$\therefore x^{2}-2x + 1+z^{2}+4z + 4 = 0$
$\therefore (x - 1)^{2}+(z + 2)^{2}= 0$
$\therefore x - 1 = 0$，$z + 2 = 0$
解得$x = 1$，$z=-2$
把$x = 1$代入$x + y = 2$得$y = 1$
所以$xyz=1×1×(-2)= - 2$
综上，答案依次为：(1)$a = 3$，$b=-1$；(2)7；(3)$-2$。