Lines are drawn parallel to the line $4x-3y+2=0$, at a distance $3/5$ from the origin. Then which one of the following points lies on any of these lines?

The given line is $4x-3y+2=0$ The straight line parallel to the given line is 

$4x-3y+c=0$ Distance of this line from origin is 3/5.

$\therefore \left|\frac{c}{\sqrt{16+9}}\right|=\frac{3}{5}\Rightarrow \left|\frac{c}{5}\right|=\frac{3}{5}\Rightarrow c=\pm 3$

So, the parallel lines are $4x-3y+3=0$ or $4x-3y-3=0$

Now, Point $\left(-\frac{1}{4},\frac{2}{3}\right)$ given in option (c) satisfies the equation, $4x-3y+3=0.$ So, $\left(-\frac{1}{4},\frac{2}{3}\right)$ lies on the line $4x-3y+3=0$