The given infinite grid consists of hexagonal cells of six resistors each of resistance R. Theo R₁₂ :

Here we will apply superposition principle.
i. Let us feed a current I to point I, but do not draw the current from 2 and allow the current to spread to infinity. Then current in branch 12 will be I/3.
ii. Secondly, feed current I to the current at take it out from 2, now again current through branch 12 will be I/3.
iii. Now carry out both the above steps (i) and (ii) simultaneously. Then current in branch 12 will be sum of above two currents
$\text{ i.e., }\frac{\mathrm{I}}{3}+\frac{\mathrm{I}}{3}=\frac{2\mathrm{I}}{3}$
Resistance between 1 and 2 is R. Let remaining resistance is R,. Then
$\begin{array}{l}{\mathrm{R}}_1\frac{\mathrm{I}}{3}=\mathrm{R}\frac{2\mathrm{I}}{3}\text{ or }{\mathrm{R}}_1=2\mathrm{R}\\ {\mathrm{R}}_{\mathrm{eq}}=\frac{{\mathrm{RR}}_1}{\mathrm{R}+{\mathrm{R}}_1}=\frac{\mathrm{R}2\mathrm{R}}{\mathrm{R}+2\mathrm{R}}=\frac{2\mathrm{R}}{3}\end{array}$