Question

If the equation of the plane containing the lines $x-y-z-4=0,x+y+2z-4=0$ and parallel to the line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=2$ is $x+Ay+Bz+C=0$, then the value of $|A+B+C|$ is

Answer 1

Equation of plane containing lines $x-y-z-4=0$ and $x+y+2z-4=0$ is
$P_1+\lambda P_2=0\Rightarrow (x-y-z-4)+\lambda (x+y+2z-4)=0$
Normal vector to the plane is
$\overrightarrow{n}=(1+\lambda )\hat{i}+(-1+\lambda )\hat{j}+(-1+2\lambda )\hat{k}$

Finding the direction ratio of the line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=2$
Putting $x=0$, we get
$3y+z=1,3y+2z=2\Rightarrow z=1,y=0$
So, the point of intersection is $(0,0,1)$
Now, putting $z=0$, we get
$2x+3y=1,x+3y=2\Rightarrow x=-1,y=1$
So, the point of intersection is $(-1,1,0)$
Therefore, the DR's of the line of intersection $=(1,-1,1)$
It is perpendicular to the normal vector, so
$(1+\lambda )\hat{i}+(-1+\lambda )\hat{j}+(-1+2\lambda )\hat{k}\cdot (\hat{i}-\hat{j}+\hat{k})=0\Rightarrow 1+\lambda +1-\lambda -1+2\lambda =0\Rightarrow \lambda =-\frac{1}{2}$

Therefore, the equation of plane is
$x-3y-4z-4=0$
Comparing with $x+Ay+Bz+C=0$, we get
$\frac{1}{1}=\frac{-3}{A}=\frac{-4}{B}=\frac{-4}{C}\Rightarrow A=-3,B=-4,C=-4\therefore |A+B+C|=11$