Let the equation of the plane containing the line `x-y - z -4=0=x+y+ 2z-4` and is parallel to the line of intersection of the planes `2x + 3y +z=1` and `x + 3y + 2z = 2` be `x + Ay + Bz + C=0` Compute the value of `|A+B+C|`.

Options

A. 4
B. 2
C. 6
D. 8

Answer

C

Explanation

A plane containing the line of intersection of the given planes is
`" "x-y-z-4+lamda(x+y+2z-4)=0`
i.e., `(lamda+1)x+ (lamda-1)y+ (2lamda-1)z-4(lamda+1)=0`
vector normal to it
`" "V=(lamda+1)hati+ (lamda-1)hatj+ (2lamda-1)hatkS" "` (i)
Now the vector along the line of intersection of the planes `2x+3y+z-1=0 and x+3y+2z-2=0` is given by
`" "vecn= |{:(hati,,hatj,,hatk), (2,,3,,1), (1,,3,,2):}|= 3(hati-hatj+hatk)`
As `vecn` is parallel to the plane (i), we have
`" "vecn*vecV=0`
`" "(lamda+1)- (lamda-1)+ (2lamda-1)=0`
`" "2+2lamda-1=0 or lamda = (-1)/(2)`
Hence, the required plane is
`" "(x)/(2)-(3y)/(2)-2z-2=0`
`" "x-3y-4z-4=0`
Hence, `|A+B+C|=6`