Given,

Binomial expression,

${\left(\sqrt{2^{{\log }_2\left(10-3^x\right)}}+\sqrt[5]{2^{(x-2){\log }_{2}3}}\right)}^{\mathrm{m}}$

${\left(\sqrt{\left(10-3^x\right)}+\sqrt[5]{3^{(x-2)}}\right)}^{\mathrm{m}}$

Now, $T_6={}^{m}C_5{\left(10-3^x\right)}^{\frac{m-5}{2}}·\left(3^{x-2}\right)=21 \dots \left(1\right)$

Also given,

${}^{m}C_1,{}^{m}C_2,{}^{m}C_3$ are in A.P.

So, $2·{}^{m}C_2={}^{m}C_1+{}^{m}C_3$

$\Rightarrow 2\times \frac{m!}{2!\left(m-2\right)!}=m+\frac{m!}{3!\left(m-3\right)!}$

Solving for $m$, we get $m=2 , 7$ and $m=2$ (rejected), so  $m=7$

Put in equation $\left(1\right)$

$21·\left(10-3^x\right)\frac{3^x}{9}=21$

$\Rightarrow \left(10-3^x\right)3^x=9\times 1$

$\Rightarrow 3^x=3^0,3^2$

$\Rightarrow x=0, 2$

Sum of the squares of all possible values of $x=4$.