If the plane 4(x−1)+k(y−2)+8(z−5)=0 contains the line x−12=y−24=z−53, then  k is

If a plane contains the line, then its normal and line are perpendicular ∴  4(2)+k(4)+8(3)=0(i.e., al+bm+cn=0)  8+4k+24=0⇒k=−8