The given equation of lines are 9x + 6y – 7 = 0 ⇒ 3 + 2y - $\frac{7}{3}$ = 0 ... (i) and 3x + 2y + 6 = 0 ... (ii) Let the equation of the line mid-way between the parallel lines (i) and (ii) be 3x + 2y + λ = 0 ... (iii) Then, distance between (i) and (iii) = distance between (ii) and (iii) ⇒ $\frac{ \left| \lambda + \frac{7}{3} \right| }{ \sqrt{9+4} }$ = $\frac{ \left| \lambda - 6 \right| }{ \sqrt{9-4} }$ ⇒ $\left| \lambda + \frac{7}{3} \right|$ = |λ - 6| ⇒ λ + $\frac{7}{3}$ = ± (λ - 6) ⇒ λ + $\frac{7}{3}$ = λ - 6 (not possible) λ + $\frac{7}{3}$ = - λ + 6 ⇒ 2λ = 6 - $\frac{7}{3}$ = $\frac{11}{3}$ ⇒ λ = $\frac{11}{6}$ Hence, the equation of the required line is 3x + 2y + $\frac{11}{6}$ = 0 i.e., 18x + 12y + 11 = 0.