We have, 3lm−4ln+mn=0 . . . (i) and l+2m+3n=0 . . . (ii) From Eq. (ii), l=−(2m+3n) Using in Eq. (i), we get −3(2m+3n)m+4(2m+3n)n+mn=0 ⇒−6m2−9mn+8mn+12n2+mn=0 ⇒−6m2+12n2=0 Now, m2=2n2⇒m=±√2n Now, beginaligned⇒=√2n ⇒l=−(2√2n+3n)=−(2√2+3)n ∴l:m:n=−(3+2√2)n:√2n:n =−(3+2√2):√2:1 Also, m=−√2n⇒l=−(−2√2+3)n l:m:n=−(3−2√2)n:−√2n:n =−(3−2√2):−√2:1endaligned If θ is the angle between the lines, then cosθ=