AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

Mid point of the line segment joining the points P(3,2,4)‌and‌Q(−1,0,−2) is R(1,1,1) and direction of the line segment PQ is 4,2,6 so, direction ratio of normal to the plane is,
(a,b,c)=(4,2,6)
So, equation of plane will be,
4x+2y+6z+d=0
Since the above plane bisect the line segment joining PQ so,
d=−12
Therefore,
ac+bd=0