If P₁, P₂ denote the lengths of the perpendiculars from the point (2,3) on the lines given by $15x^2+31xy+14y^2=0$ if $P_1>P_2$ then  $P_1^2+\frac{1}{74}-P_2^2+\frac{1}{13}=$ _____

$$\begin{gathered} 15x^2+31xy+14y^2=0\text{ represent the lines }5x+7y=0 \to \left(1\right) \\ and 3x+2y=0\to \left(2\right) \end{gathered}$$

$$\begin{gathered} P_1= perpendicular dis\tan ce from (2,3) to \left(1\right) \\ =31/\sqrt{74} \\ {P}_2=perpendicular dis\tan ce from (2,3) to \left(2\right) \\ =12/\sqrt{13} \\ where P_1>P_2 \\ Now {P^2}_1+\frac{1}{74}-{P_2}^2+\frac{1}{13}=2 \end{gathered}$$