Question

Let the equation of the plane containing the 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 $|A+B+C|$ is

• A. 11.0
• B. 11
• C. 11.00

Answer 1

Answer: (A), (B), (C)

Plane containing the line of intersection of the given planes is
$(x-y-z-4)+\lambda (x+y+2z-4)=0$
$\Rightarrow (\lambda +1)x+(\lambda -1)y+(2\lambda -1)z-4(\lambda +1)=0\cdots \left(1\right)$
Let $p,q,r$ be the direction ratios of line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=2.$
Then, $2p+3q+r=0\cdots \left(2\right)$
and $p+3q+2r=0\cdots \left(3\right)$
From equations $\left(2\right)$ and $\left(3\right),$ we get
$\frac{p}{6-3}=\frac{q}{1-4}=\frac{r}{6-3}$
$\Rightarrow \frac{p}{1}=\frac{q}{-1}=\frac{r}{1}$
Since plane $\left(1\right)$ is parallel to this line
$\therefore (\lambda +1)\left(1\right)+(\lambda -1)(-1)+(2\lambda -1)\left(1\right)=0$
$\Rightarrow \lambda +1-\lambda +1+2\lambda -1=0$
$\Rightarrow \lambda =-\frac{1}{2}$
From $\left(1\right),$ we have
$\frac{x}{2}-\frac{3y}{2}-2z-2=0$
$\Rightarrow x-3y-4z-4=0$
Comparing it with $x+Ay+Bz+C=0$
$\therefore A=-3,B=-4,C=-4$
Hence, the value of $|A+B+C|$ is $11.$