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

If a plane contains the line, then its normal and line are perpendicular
$$\begin{gathered} \therefore 4\left(2\right)+\mathrm{k}\left(4\right)+8\left(3\right)=0\Rightarrow (\text{ i.e., }\mathrm{al}+\mathrm{bm}+\mathrm{cn}=0) \\ \Rightarrow 32+4k=0 \\ \Rightarrow 4k=-32 \\ \Rightarrow k=-8 \end{gathered}$$