Solution:

From the values of op voltage and output current it is clear that this is a low voltage Low current Regulator [LVLC]

To calculate $R_1$ and $R_2$ :-

Assume that current through $R_1$ and $R_2$ to be equal to $1 \mathrm{~mA}$

$ \begin{aligned} \therefore R_2 &=\frac{V}{I}=\frac{5 \mathrm{~V}}{1 \mathrm{~mA}}=5 \mathrm{k} \Omega \\ \therefore R_2 &=5 \mathrm{k} \Omega \end{aligned} $

$ \begin{aligned}\\ \text { Voltage across } R_1 &=\text { Vref }-V R_2\\ \\ &=\text { vref }-V 0 \\\\ &=7-5 \\\\ &=2 \mathrm{Volts} . \\\\ \therefore R_1=\frac{[V r e f-V 0]{ }}{I} &=\frac{2}{1 \mathrm{~mA}}=2 \mathrm{k} \Omega \\\\ \therefore R_1=2 \mathrm{k} \Omega\\ \end{aligned}\\ $

To calculate $R_3$ :

$ \begin{aligned} &R_3=R_1 11 R_2 \\ &R_3=[5 \mathrm{k} \Omega 112 \mathrm{k} \Omega] . \\ &R_3=1.4 \mathrm{k} \Omega \end{aligned} $

To calculate $R_sc$:-

$ R_sc=\frac{V s e n s e}{I s c} $

$ \begin{aligned}\\ &R S C=\frac{0.6}{75 m A} \\\\ &R_{S C}=8 \Omega\\ \end{aligned}\\ $

To select $c_1$ :-

$\bar{C}_1$ is an additional filter capacitor so let us select its value to be $4.7 \mu \mathrm{F} / 10 \mathrm{~V}$