The given equation of lines are 2x – 3y + 4 = 0 ...(i) and 3x + 4y – 5 = 0 ... (ii) On solving (i) and (ii), we get x = $-\frac{1}{17}$ , y = $\frac{22}{17}$ ∴ Required point is $\left(-\frac{1}{17},\frac{22}{17}\right)$ Now slope of 6x – 7y + 8 = 0 is $\frac{6}{7}$ ∴ Required slope is $-\frac{7}{6}$ Therefore required equation of a line is $\left(y-\frac{22}{17}\right)$ = $-\frac{7}{6}\left(x+\frac{1}{17}\right)$ ⇒ (17y – 22) 6 = – 7(17x + 1) ⇒ 102y – 132 = – 119x – 7 ⇒ 119x + 102y – 125 = 0 ⇒ 119x + 102y = 125.