If the value of $(a+b-2)^2+(b+c-5)^2+(c+a-5)^2=0$, then the value of $\sqrt{(b+c)^a+(c+a)^b-1}$ is:

3

Given,

 (a + b - 2)² + (b + c - 5)² + (c + a - 5)² = 0

So,

(a + b - 2) = 0

= a + b = 2     ---- (1)

(b + c - 5) = 0

= b + c = 5     ---- (2)

(c + a - 5) = 0

= c + a = 5     ---- (3)

Solving, (1), (2) and (3), we get

a = 1, b = 1 and c = 4

Then, $\sqrt{(b+c)^a+(c+a)^b-1}$ = $\sqrt{(1+4)^1+(4+1)^1-1}$ = 3