The vector equation of the line passing through the point (-1, -1 ,2) and parallel to the line 2x - 2= 3y +1 = 6z -2 is

Explanation

Equation of the line is
`2x-2=3y+1=6z-2`
Dividing throughout by 6, we get
`(2(x-))/(6)=(3(y+(1)/(3)))/(6)=(6(z-(2)/(6)))/(6)`
`:.(x-1)/(3)=(y+(1)/(3))/(2)=(z-(1)/(3))/(1)`
`:.` Direction ratio of the line are 3, 2, 1.

Let `vec(b)` be the vector parallel to this line
`:.vec(b)=-hati+2hatj+hatk" "......(i)`
Let `vec(a)` be the position vector of the point (-1,-1,2)
`:.vec(a)=-hati-hatj+2hatk" "......(ii)`
The vector equation of a line passing through the piont (-1, -1, 2) with position vector `vec(a)` and parallel to `vec(b)" is "vec(r)=vec(a)+lamdavec(b)`
`:.` Vector equation of the required line is
`vec(r)=(-hati-hatj+2hatk)+lamda(3hatj+2hatj+hatk)`
.......(using (i) and (ii)