Correct Answer - B
Lines is statement -1 pass through (1,0,-1) and (2,-1,0) respectively.
We have
`|(2-1,-1-0 ,0+1),(1,-1,1),(1,2,3)|=5+2+3=0`
So the given lines are coplanar.
The equation of the plane containing them is
`|(x-1,y,z+1),(1,-1,1),(1,2,3)|=0`
`implies -5x+5-2y+3z+3=0implies5x+2y-3z-8=0`
So statement -1 is true.
A vector parallel to the given line is `vecb=hati+2hatj+3hatk` and vectors normals to given planes are `vecn_(1)=3hati+6hatj+9hatk` and `vecn_(2)=hati+hatj-hatk` respectively.
Clearly` vecn_(1)=3vecb` and `vecb.vecn=0`
So, give line is perpendicular to `3x+6y+9z-8=0` and parallel to `x+y-z=0`
Hece statement -2 is true.