Let the sixth term in the binomial expansion of $$\left(\sqrt{2^{\log_2(10-3^x)}} + \sqrt[5]{2^{(x-2)\log_2 3}}\right)^m$$ powers of $$2^{(x-2)\log_2 3}$$ be 21. If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of $$x$$ is _____.

$$^mC_1, \quad ^mC_2, \quad ^mC_3 \text{ form an A.P.}$$

$$2(^mC_2) = ^mC_1 + ^mC_3$$

$$2 \cdot \frac{m(m-1)}{2} = m + \frac{m(m-1)(m-2)}{6}$$

$$m-1 = 1 + \frac{(m-1)(m-2)}{6} \implies 6m - 6 = 6 + m^2 - 3m + 2$$

$$m^2 - 9m + 14 = 0 \implies (m-2)(m-7) = 0 \implies m = 7 \quad (\text{since } m \ge 5 \text{ for } T_6)$$

Simplifying terms of the binomial expression $$(a + b)^7$$:

$$a = 2^{\frac{1}{2}\log_2(10-3^x)} = (10-3^x)^{1/2}$$

$$b = 5^{(x-2)\log_5 2} = 2^{x-2}$$

$$T_6 = ^7C_5 \cdot a^2 \cdot b^5 = 21$$

$$21 \cdot \left((10-3^x)^{1/2}\right)^2 \cdot \left(2^{x-2}\right)^5 = 21$$

$$(10-3^x) \cdot 2^{5x-10} = 1$$

$$10-3^x = 2^{10-5x}$$

$$\text{By inspection, at } x = 2:$$

$$10 - 3^2 = 2^{10-10} \implies 1 = 1 \quad (\text{Valid})$$

$$\text{For real domain condition: } 10 - 3^x > 0 \implies 3^x < 10$$

$$x = 2 \text{ is the only real solution.}$$

$$\sum x^2 = 2^2 = 4$$