If the equation of the plane bisecting the line segment joining the points $P(3,2,4)$ and $Q(-1,0,-2)$ and perpendicular to $P Q$ is $a x+b y+c z+d=0$, then $a c+b d$

Mid-point of line segment joining the points $P(3,2,4)$ and $Q(-1,0,-2)$ is $R(1,1,1)$ and direction ratios of line segment $P Q$ is $4,2,6$, so direction ratios of normal to the plane is $$ \langle a, b, c\rangle=\langle 4,2,6\rangle . $$ So, equation of plane will be $$ 4 x+2 y+6 z+d=0 $$ Since, plane (i) bisect the line segment joining $P Q$. So, $\quad d=-12$ Therefore, $\quad a c+b d=0$.