Points $A$ lies on $L _{2}$

$A \left(\alpha, 4+\frac{4}{3} \alpha\right)$

Points $B$ lies on $L _{1}$

$B \left(\beta, 2-\frac{2}{5} \beta\right)$

Points $P$ divides $AB$ internally in the ratio $1: 3$

$\Rightarrow P(2,3)=P\left(\frac{3 \alpha+\beta}{4}, \frac{3\left(4+\frac{4}{3} \alpha\right)+1\left(2-\frac{2}{5} \beta\right)}{4}\right)$

$\Rightarrow \alpha=\frac{3}{13}, \beta=\frac{95}{13}$

Point $A \left(\frac{3}{13}, \frac{56}{13}\right), B \left(\frac{95}{13},-\frac{12}{13}\right)$

Vertex $C$ of triangle is the point of intersection of $L _{1} and L _{2}$

$\Rightarrow C \left(-\frac{15}{13}, \frac{32}{13}\right)$

area $\triangle ABC =\frac{1}{2}\left\|\begin{array}{ccc}\frac{3}{13} & \frac{56}{13} & 1 \\ \frac{95}{13} & -\frac{12}{13} & 1 \\ -\frac{15}{13} & \frac{32}{13} & 1\end{array}\right\|$

$\frac{1}{2 \times 13^{3}}\left\|\begin{array}{ccc} 3 & 56 & 13 \\ 95 & -12 & 13 \\ -15 & 32 & 13 \end{array}\right\|$

area $\triangle ABC =\frac{132}{13}$ sq. units.