It is known that the lengths of three sides of △ ABC are a, B, C, and the perimeter is 6, and a ^ 2 + B ^ 2 + C ^ 2 = AB + BC + Ca, then the lengths of three sides of △ ABC are?

a^2+b^2+c^2=ab+bc+ca
=>2a²+2b²+2c²-2ab-2ac-2ca=0
=>(a-b)²+(b-c)²+(c-a)²=0
=>a=b,b=c,c=a
=>a=b=c,a+b+c=6
=>a=b=c=2

Given that ABC is the trilateral length BC of △ satisfying (b-2) | C-3 | = 0, and a is the solution of the equation | x-4 | = 2, find the perimeter of △ ABC and judge its shape

（b-2）|c-3|=0
b=2 c=3
A is the equation | x-4 | = 2, x = 6, x = 2
b+c>a a=2
The circumference of △ ABC is a + B + C = 2 + 2 + 3 = 7 isosceles triangle

It is known that ABC is the length of three sides of triangle, and a ^ 2 + B ^ 2 + C ^ 2-6a-8b-10c + 50 = 0, please judge its perimeter

a^2+b^2+c^2-6a-8b-10c+50=0
(a-3)^2+(b-4)^2+(c-5)^2=0
So:
a-3=0,b-4=0,c-5=0
a=3,b=4,c=5

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