...(ii) Let a point (2r + 1, 3r - 1, 4r + 1) be on the line Eq. (i). If this is an intersection point of both the lines, then it will lie on Eq. (ii), also ∴
2r+1−3
1
=
3r−1−k
2
=
4r+1
1
(iii) Taking first and third part of eq. (iii), we get 2r−2=4r+1 ⇒ r=−
3
2
Taking second and third part of eq. (iii), we get 3r−1−k=8r+2 ⇒ 3r−1−k−8r−2=0 ⇒ k=−5r−3 ⇒ k=−5(−