If  $a,b,c$  are positive real numbers such that  $a^{{\log }_{3}7}=27;b^{{\log }_{7}11}=49$  and  $c^{{\log }_{11}25}=\sqrt{11}$. The unit digit in the value of  $(a^{{\left({\log }_{3}7\right)}^2}+b^{{\left({\log }_{7}11\right)}^2}+c^{{\left({\log }_{11}25\right)}^2})$  equals 

 $$\begin{gathered} T_1=a^{\left({\log }_{3}7\right)\left({\log }_{3}7\right)}={27}^{{\log }_{3}7}=3^{3{\log }_{3}7}=343 \\ {T}_2=b^{\left({\log }_{7}11\right)\left({\log }_{7}11\right)}={49}^{{\log }_{7}11}=7^{2{\log }_{7}11}=121 \\ {T}_3=c^{\left({\log }_{11}25\right)\left({\log }_{11}25\right)}={\left(\sqrt{11}\right)}^{{\log }_{11}25}={11}^{\frac{1}{2}{\log }_{11}25}=5 \\ \therefore sum=T_1+T_2+T_3=343+121+5=469 \end{gathered}$$