First line is (ϕ+α,2ϕ+1,3ϕ+1) and second line is (qβ+4,3q+6,3q+7). For intersection ϕ+α=qβ+4 .....(i) 2ϕ+1=3q+6 ......(ii) 3ϕ+1=3q+7........(iii) for (ii) & (iii) ϕ=1,q=−1 So, from (i) α+β=3 Now, point of intersection is (α+1,3,4) It lies on the plane. Hence, α=5&β=−2