$I=1\, \mathrm{A}$        .......$(i)$

When the key between the terminals $1$ and $2$ is plugged in, then

Potential difference across $R=I R=k l_{1}$       ......$(ii)$

where $k$ is the potential gradient across the potentiometer wire

When the key between the terminals $1$ and $3$ is plugged in, then

Potential difference across $(R+X)=I(R+X)=k l_{2}$      ....$(iii)$

From equation $(ii),$ we get

$R=\frac{k l_{1}}{I}=\frac{k l_{1}}{1}=k l_{1} \Omega$      .......$(iv)$

From equation $(iii),$ we get

$R + X = \frac{{k{l_2}}}{I} = \frac{{k{l_2}}}{1} = k{l_2}\,\Omega \quad {\rm{ (Using }}({\rm{i}}))$

$X = k{l_2} - R$

$ = k{l_2} - k{l_1}{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,{\rm{(Using}}\left( {iv} \right){\rm{)}}$

$=k\left(l_{2}-l_{1}\right) \,\Omega$