Question

Let the equation of the plane containing line $x-y-z-4=0, x+y+2 z-4$ and parallel to the line of intersection of the planes $2 x+3 y+z=1$ and $x+3 y+2 z=2$ be $x+A y+B z+C=0$. The find the value of $| A + B + C -4|$.

Answer 1

A plane containing the line of intersection of the given planes is.
$x-y-z-4+\lambda(x+y+2 z-4)=0$
i.e., $(\lambda+1) x +(\lambda-1) y +(2 \lambda-1) z -4(\lambda+1)=0$
vector normal to it
$V =(\lambda+1) \hat{ i }+(\lambda-1) \hat{ j }+(2 \lambda-1) \hat{ k } S...$(i)
Now the vector along the line of intersection of the planes
$2 x+3 y+z-1=0$ and $x+3 y+2 z-2=0$ is given by
$\vec{ n }=\begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{vmatrix}=3(\hat{ i }-\hat{ j }+\hat{ k })$
As $\vec{ n }$ is parallel to the plane (i), we have
$\vec{ n } \cdot \vec{ V }=0$
$(\lambda+1)-(\lambda-1)+(2 \lambda-1)=0 $
$2+2 \lambda-1=0 $ or $ \lambda=\frac{-1}{2}$
Hence, the required plane is.
$\frac{x}{2}-\frac{3 y}{2}-2 z-2=0 $
$x-3 y-4 z-4=0$
Hence, $|A+B+C|=6$