If the $6^{th}$ term in the expansion of the binomial $[\sqrt{2^{\log(10 - 3^x)}} + \sqrt[5]{2^{(x - 2)\log 3}}]^m$ is equal to $21$ and it is known that the binomial coefficients of the $2^{nd}$,$3^{rd}$ and $4^{th}$ terms in the expansion represent respectively the first,third and fifth terms of an $A.P.$ (the symbol $\log$ stands for logarithm to the base $10$),then $x = $
- A$0$
- B$1$
- C$2$
- D$0$ or $2$
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