So, (1,−4,−2) lies on plane. 1−4m−2n=0...(i) And (−1,2,3) will be perpendicular to (I,m,n). −1+2m+3n=0....(ii) Adding Eqs. (i) and (ii), −2m+n=0 n=2m l−4m−4m=0 l=8m So, l=8m and n=2m Plane ⇒8x+y+2z=0 Now, A(−3,−6,1) and B(2,4,−3) Plane P divides AB in the ratio of k:1. Let plane P intersect the line AB at point O. So, O=(
2k−3
k+1
,
4k−6
k+1
,
−3k+1
k+1
) And O lies on plane P, So, 8(2k−3)+(4k−6)+2(−3k+1)=0 ⇒14k−28=0 ∴k=2