Correct Answer - D
The equation of the lines joining `6 vec a-4 vecb+vecc, -4 vecc`
`-veca -2vecb-3, vecc, veca+2 vecb-5 vecc` are respectively
`vecr=6 vec a- 4 vecb+ 4 vecc+ m(-6 vec a+4 veca+4 vecb-8 vecc)" "...(i)`
and `vecr=- veca-2 vecb- 3 vecc+ n(2 veca+ 4 vecb- 2 vecc) " "....(ii)`
For the point of intersection the Eqs. (i) nd (ii) should give the same value of `vec.` Hence, eqauating the coeeficients of vectors `veca, vecb and vecc` in two expression for `vecr`, we get
`{:(6m,2n,=7,.......(i)),(2m,-2n,=1,.....(iv)),( 8m,-2n,=7,.....(v)):}`
On solving Eqs (iii) and (iv), we get `m=1,n=(1)/(2)`. These values of m and n also satify the Eq (v)
`:.` The lines intersect . Puting the value of m in Eq (i). we get fthe position vector of the point