If the plane $4(x-1)+k(y-2)+8(z-5)=0$ contains the line $\frac{x-1}{2}=\frac{y-2}{4}=\frac{z-5}{3}$, then  $k$ is

If a plane contains the line, then its normal and line are perpendicular 
$\therefore 4\left(2\right)+k\left(4\right)+8\left(3\right)=0$
$(i.e.,al+bm+cn=0)$
  $8+4k+24=0$$\Rightarrow k=-8$