Let the sixth term in the binomial expansion of $${\left( {\sqrt {{2^{{{\log }_2}\left( {10 - {3^x}} \right)}}} + \root 5 \of {{2^{(x - 2){{\log }_2}3}}} } \right)^m}$$ in the increasing 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 __________.

${ }^m C_1,{ }^m C_2,{ }^m C_3$ are first, third and fifth term of $A P$ <br/><br/>$$ \begin{aligned} \therefore \quad & a={ }^m C_1 \\\\ & a+2 d={ }^m C_2 \\\\ & a+4 d={ }^m C_3 \\\\ \therefore \quad & 2{ }^m C_2-{ }^m C_3=m \\\\ \Rightarrow & m=7 \text { or } m=2 \\\\ \because & m=2 \text { is not possible } \\\\ \therefore & m=7 \end{aligned} $$ <br/><br/>$\mathrm{T}_6={ }^{\mathrm{m}} \mathrm{C}_5\left(10-3^{\mathrm{x}}\right)^{\frac{\mathrm{m}-5}{2}} \cdot\left(3^{\mathrm{x}-2}\right)=21$ <br/><br/>Putting value of m = 7, we get <br/><br/>$\begin{aligned} & T_{5+1}={ }^7 C_5\left(10-3^x\right)^{\frac{7-5}{2}} 3^{x-2}=21 \\\\ & \Rightarrow \frac{10.3^x-\left(3^x\right)^2}{3^2}=1\end{aligned}$ <br/><br/>$$ \begin{aligned} & \Rightarrow \left(3^x\right)^2-10 \cdot 3^x+9=0 \\\\ & \Rightarrow 3^x=9,1 \\\\ & \Rightarrow x=0,2 \end{aligned} $$ <br/><br/>Sum of squares of values of x = 0² + 2² = 4