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In how many ways three letters a,b,c can...

In how many ways three letters a,b,c can be arranged without repetition ?

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The correct Answer is:
C
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`
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