Q107 / 170 · Manipal (MET) 2016 · Mathematics
The equation of the plane passing through the line of intersection of the planes $x+y+z=6$ and $2 x+3 y+4 z+5=0$ and perpendicular to the plane $4 x+5 y-3 z=8$ is
- A$x+7 y+13 z-96=0$
- Correct answerB$x+7 y+13 z+96=0$
- C$x+7 y-13 z-96=0$
- D$x-7 y+13 z+96=0$
Solution
The intersection of two planes is
$\begin{aligned} & (x+y+z-6)+\lambda(2 x+3 y+4 z+5)=0 \\ & \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1+4 \lambda) z+(-6+5 \lambda)=0\ldots(i)\end{aligned}$
Since, this plane is perpendicular to the plane
$4 x+5 y-3 z-8=0$
$\therefore \quad(1+2 \lambda) 4+(1+3 \lambda) 5+(1+4 \lambda)(-3)=0$
$\Rightarrow \quad \lambda=\frac{-6}{11}$
On putting the value of $\lambda$ in Eq. (i), we get
$\left(-\frac{1}{11}\right) x+\left(-\frac{7}{11}\right) y+\left(-\frac{13}{11}\right) z+\left(-\frac{96}{11}\right)=0$
$\Rightarrow \quad x+7 y+13 z+96=0$
Hence, the correct answer is option: 'B'.