Let $t$ be the normal time he takes and let $S$ be the distance.

$$\begin{equation} \frac{S}{\frac{S}{t} + 1} = \frac{3t}{4} \\\implies \frac{S}{S+t} =\frac{3}{4} \\\implies S = 3t \label{670eq1} \end{equation}$$

$$\begin{equation} \frac{S}{\frac{S}{t} -1} = t + 2 \\\implies \frac{St}{S-t} = t + 2 \label{670eq2} \end{equation}$$

From $\ref{670eq1}$ and $\ref{670eq2}$,

$$\frac{3t^2}{2t} = t+2 \\\implies t = 4\\\implies S = 12kms$$.