Let a triangle be bounded by the lines L₁ : 2x + 5y = 10; L₂ : $-$4x + 3y = 12 and the line L₃, which passes through the point P(2, 3), intersects L₂ at A and L₁ at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :

Let $A=(a,b)\in L_2\Rightarrow -4a+3b=12.$ Since $AP:PB=1:3$, by section formula $P=\dfrac{B+3A}{4}\Rightarrow B=4P-3A=(8-3a,\;12-3b).$ Because $B\in L_1$, $2(8-3a)+5(12-3b)=10\Rightarrow 2a+5b=22.$ Solve \[ \begin{cases} 2a+5b=22,\\ -4a+3b=12 \end{cases} \Rightarrow a=\dfrac{3}{13},\quad b=\dfrac{56}{13}. \] Thus \[ A=\left(\dfrac{3}{13},\dfrac{56}{13}\right),\quad B=\left(\dfrac{95}{13},-\dfrac{12}{13}\right). \] Intersection $C=L_1\cap L_2$: \[ \begin{cases} 2x+5y=10,\\ -4x+3y=12 \end{cases} \Rightarrow C=\left(-\dfrac{15}{13},\dfrac{32}{13}\right). \] Area \[ \Delta=\frac12\left| \begin{vmatrix} x_A&y_A&1\\ x_B&y_B&1\\ x_C&y_C&1 \end{vmatrix}\right| =\frac12\left|(B-A)\times(C-A)\right| =\frac12\left|(92)(-24)-(-68)(-18)\right| =\frac{1716}{169} =\boxed{\dfrac{132}{13}}. \] Answer: $\boxed{\dfrac{132}{13}}$.