The line x + 3y - 2 = 0 bisects the angle between a pair of straight lines of which one has equation x - 7y + 5 = 0.The equation of the other line is 

 The family of line through the given lines is 

$$\begin{gathered} L\equiv x-7y+5+\lambda (x+3y-2)=0 \\ =x(1+\lambda )+y(-7+3\lambda )+5-2\lambda -----\left(1\right) \end{gathered}$$

let  (2, 0) be a point on x+3y-2=0 which is equidistant from  x - 7y + 5 = 0 and the line L = 0

Therefore,

$\left|\frac{2-0+5}{\sqrt{1^2+7^2}}\right|=\left|\frac{2(1+\lambda )+0(-7+3\lambda )+5-2\lambda }{\sqrt{{\left(1+\lambda \right)}^2+{(-7+3\lambda )}^2}}\right|$

$\left|\frac{2+5}{\sqrt{50}}\right|=\left|\frac{2+2\lambda +5-2\lambda }{\sqrt{(1+\lambda {)}^2+(3\lambda -7{)}^2}}\right|$

$\begin{array}{l}\text{ or }10{\lambda }^2-40\lambda =0\\ \text{ i.e., }\lambda =4\text{ or }0\\ \text{ Hence, }L=0,\lambda =4.\end{array}$

$\text{ Therefore, the required line is }5x+5y-3=0\text{ . }$