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 $\overline{PQ}$ is $ax+by+cz+d=0$ then $ac+bd=$

P = (3, 2, 4), Q = (-1, 0, -2)

 Let ‘R’ is midpoint of $\overline{PQ}$

$R=\frac{P+Q}{2}=[1,1,1]$

D.r’s of $\overline{PQ}$ are $[x_2-x_1,y_2-y_1,z_2-z_1]$ 

= [-4, -2, -6]

Eq. Of the plane $\text{'}{\perp }^{er}\text{'}$ to $\overline{PQ}$ is

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$

$-4(x-1)-2(y-1)-6(z-1)=0$

$\Rightarrow 4x+2y+6z-12=0$

$a=4|b=2|c=6|d=-12$

$\to ac+bd=24-24=0$