Question

Let the 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 $2x+3y+z=1$ and $x+3y+2z=2$ be $x+Ay+Bz+C=0$. Then the value of $|A+B+C-4|$ is

Answer 1

A plane containing the line of intersection of the given planes is $x-y-z-4+\lambda (x+y+2z-4)=0$ i.e., $(\lambda +1)x+(\lambda -1)y+(2\lambda -1)z-4(\lambda +1)=0$ vector normal to it.
$V=(\lambda +1)\hat{i}+(\lambda -1)\hat{j}+(2\lambda -1)\hat{k}\dots \left(1\right)$
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
$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 1\\ 1& 3& 2\end{array}\right|=3(\hat{i}-\hat{j}+\hat{k})$

As $\overrightarrow{n}$ is parallel to the plane $\left(1\right)$, we have
$\overrightarrow{n}.\overrightarrow{V}=0$
$\Rightarrow (\lambda +1)-(\lambda -1)+(2\lambda -1)=0$
$\Rightarrow \lambda =\frac{-1}{2}$

Hence, the required plane is
$\frac{x}{2}-\frac{3y}{2}-2z-2=0$
$x-3y-4z-4=0$
Hence, $|A+B+C|=6$