If the set of all values of x for which the expansion of  ${\left(7-5x\right)}^{-2/3}$ valid is equal to $\left(-a,a\right)$  then $5a+7$  =

$(7-5x{)}^{\frac{-2}{3}}=7^{\frac{-2}{3}}{(1-\frac{5}{7}x)}^{\frac{-2}{3}}$
$\begin{array}{rl} & \text{Now the expansion}{(1-\frac{5}{7}x)}^{\frac{-2}{3}}\text{is valid if }\left|\frac{5}{7}x\right|<1\\ & \end{array}$
$\begin{array}{rl} & \Rightarrow \left|x\right|<\frac{7}{5}\\ & \Rightarrow \frac{-7}{5}<x<\frac{7}{5}\\ & \Rightarrow a=\frac{7}{5}\\ & \text{Now }5a+7=5\left(\frac{7}{5}\right)+7=14\end{array}$