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

The equation of a line parallel to the line $ax+by+c=0$ is $ax+by+k=0.$

Thus, the equation of a line parallel to $4x-3y+2=0$ is $4x-3y+c=0.$

The distance of a line $ax+by+c=0$ from origin is $\frac{\left|c\right|}{\sqrt{a^2+b^2+c^2}}.$

Given the line $4x-3y+c=0$ is at a distance of $\frac{3}{5}$ units from origin

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

$\Rightarrow \frac{\left|c\right|}{5}=\frac{3}{5}$

$\Rightarrow c=\pm 3.$

Hence, the equation of the lines are $4x-3y+3=0$ and $4x-3y-3=0.$

Clearly point $\left(-\frac{1}{4},\frac{2}{3}\right)$ satisfy the given equation $4x-3y+3=0.$