It is given that L1 is passing through 5i+8j+11k and parallel to vector 2i+3j+4k so L1=r =5i+8j+11k+λ(2i+3j+4k)
It is given that L2 is passing through 4i+6j+8k and parallel to vector 3i+4j+5k so,
L2=r =4i+6j+8k+µ(3i+4j+5k)
Both the lines are intersecting so,
5i+8j+11k+λ(2i+3j+4k)=4i+6j+8k+µ(3i+4j+5k) i+2j+3k=(3µ−2λ)i+(4µ−3λ)j+(5µ−4λ)k Compare the coefficients on both sides, 3µ−2λ=1⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) 4µ−3λ=2⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II) From equation(I) and (II), µ=−1,λ=−2 Substitute λ=−2lnL1 then r=i+2j+3k Therefore, the point of intersection of both the lines is i+2j+3k