The given lines, r=(i+j−k)+λ(3i−j)......(i) and r=(4i−k)+µ(2i+3k).....(ii) From Eqs. (i) and (ii), (i+j−k)+λ(3i−j) =(4i−k)+µ(2i+3k) (1+3λ)i+(1−λ)j−k =(4+2µ)i+0j+(−1+3µ)k Equating the coefficient of i,j and k on both sides, 1+3λ=4+2µ 3λ−2µ=3......(iii) 1−λ=0 ⇒λ=1 −1+3µ=−1 ⇒µ=0 On putting these values in Eq. (i), r=(i+j−k)+(3i−j) r=4i+0j−k So, the intersection point is (4,0,−1)