The point of intersection of lines x+3y−1=0 and x−2y+4=0 is (−2,1). Let equation of line perpendicular to the given line is 2x−3y+λ=0. Since, it passes through (−2,1). ∴‌2(−2)−3(1)+λ=0 ⇒λ=7 ∴‌‌ Required line is ‌2x−3y+7=0