Given, 3m−4‌ln+mn=0 ...(i) l+2m+3n=0 ...(ii) ⇒ l=−2m−3n From Eq. (i), we get 3(−2m−3n)m−4(−2m−3n)n+mn=0 −6m2−9nm+8mn+12n2+mn=0 ⇒ 6m2=12n2 ⇒m2=2n2 m=±√2n ∴ (l1,m1,n1) and (l2,m2,n2) are direction ratios where, m1=√2nt and m2=√2n2 l1=−(2√2+3)n1 and l2=−(−2√2+3)n2 ∴ cos‌θ=