The given equations are 3l+m+5n=0‌‌. . .(i) and 6mn−2nl+5∕m=−0. . . (ii) Now, from Eq. (i), we get m=−3l−5n On substituting m=−3∕−5n in Eq. (ii), we get ‌‌‌6(−3l−5n)n−2nl+5l(−3l−5n)=0 ⇒‌30n2+45‌ln+15I2=0 ⇒‌2n2+3‌ln+I2=0 ⇒‌2n2+2nl+nl+I2=0 ⇒‌2n(n+l)+l(n+l)=0 ⇒‌‌‌(n+l)(2n+l)=0 ⇒‌‌ Either ‌I=−n‌ or ‌I=−2n ‌ lf ‌I=−n,‌‌ then ‌m=−2n‌ and if ‌I=−2n‌, then ‌m=n Thus, the direction ratios of two lines are proportional (−n,−2n,n) and (−2n,n,n) i.e. (−1,−2,1) and (−2,1,1), respectively. Now, let θ be the acute angle between the lines, ‌ Then, ‌cos‌θ‌‌=‌