The rate constant of a reaction is given by  $k=2.1\times {10}^{10}\cdot e^{\frac{-2700}{RT}}$
The correct options are

$\log k$ versus  $\frac{1}{T}$ will be a straight line with slope  $=\frac{-2700}{2.303R}$

$\log k$ versus  $\frac{1}{T}$, has an intercept of   $\log 2.1\times {10}^{10}$on  $\log k$ axis

$\log k$ versus $\frac{1}{T}$  has slope  $=\frac{-2700}{R}$

Half life of the reaction increases with increase of temperature.