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:

$$(a+b-2)^{2}+(b+c-5)^{2}+(c+a-5)^{2}=0$$
LHS:

$$(a+b-2)^{2}+(b+c-5)^{2}+(c+a-5)^{2}

= $$(0)^{2}+(0)^{2}+(0)^{2}
= 0
=RHS

a + b - 2 = 0

a + b = 2 ---(1)

b + c - 5 = 0

b + c = 5 ---(2)

c +a - 5 = 0

a + c = 5 ---(3)

Eq(1) + (2) + (3),

2(a + b + c) = 2 + 5 + 5

a + b + c = 6

put the value from eq(1),

2 + c = 6

c = 4

From eq(2),

b + 4 = 5

b = 1

From eq(1),

a + 1 = 2

a = 1

$$\sqrt{(b+c)^{a}+(c+a)^{b}-1}$$

=  $$\sqrt{(5)^{a}+(5)^{b}-1}$$

= $$\sqrt{(5)^{1}+(5)^{1}-1}$$

= $$\sqrt{9}$$ = 3