. . . (i) The direction ratios of the normal are (1,3,−α). The direction ratios of the line are (3,−5,2) and equation of given plane x+3y−az+β=0 . . . (ii) Four lines are perpendicular ⇒‌a1a2+b1b2+c1c2=0 ⇒‌3−15+2α=0 ⇒‌2α=−12⇒α=−6 (2,1,−2) lies on the plane, so 2+3+6(−2)+β‌‌=0⇒β=7 α⋅β‌‌=−6×7=−42