Potential across

$\mathrm{R}_1=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}$ $=V_A-V_B$    $\left(\because \mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{C}}\right)$

Potential across

$R_2=V_D-V_C$ $=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}$    $\left(\because \mathrm{V}_{\mathrm{D}}=\mathrm{V}_{\mathrm{A}}\right)$

Potential across

$R_3=V_D-V_B$ $=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}$     $\left(\because \mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{C}}\right)$

It means all the resistance are in parallel

$$ \begin{aligned} & \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3} \\\\ & \Rightarrow \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{20}+\frac{1}{20}+\frac{1}{10} \\\\ & \Rightarrow \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1+1+2}{20} \\\\ & \Rightarrow \mathrm{R_{eq}}=\frac{20}{4}=5 \mathrm{k} \Omega \end{aligned} $$