Equation of the plane containing line x−y−z−4=0=x+y+2z−4 and parallel to the line of intersection of the planes y+3y+z=1 and x+ 3y+27=2 be x+Ay+Bz+C=0 Then find the value of |A+B+C|=

A plane containing the line of intersection of the given planes is x−y−z−4+λ(x+y+2z−4)=0 i.e. (λ+1)x+(λ−1)y+(2λ−1)z−4(λ+1)=0vector normal to it V=(λ+1)i^+(λ−1)j^+(2λ−1)k^S----iNow the vector along the line of intersection of the planes 2x + 3y + z - 1 = 0 and x + 3y + 2z -2 = 0 is given byn→=i^j^k^231132=3(i^−j^+k^)As n→is parallel to the plane (i), we have n→⋅V→=0(λ+1)−(λ−1)+(2λ−1)=02+2λ−1=0  or  λ=−12Hence, the required plane isx2−3y2−2z−2=0x−3y−4z−4=0Hence,|A+B+C|=11