Correct Answer - A
The equation of the plane through the intersection of the planes `x+y+z=1` and `2x+3y-z+4=0` is
`(x+y+z-1)+lamda(2x+3y-z+4)=0`
or `(2lamda+1)x+(3lamda+1)y+(1-lamda)z+4lamda-1)=0`………..i
It is parallel to `X` -axis i.e. `x/1=y/0=z/0`
`:.1(2lamda+1)+0xx(3lamda+1)+0(1-lamda)=0implieslamda=-1/2`
Substituting `lamda=1/2` in (i) we get
`y-3z+6=0` as the equation of the required plane.
