Question

The equation of the plane passing through the line of intersection of the planes $x+y+z=6$ and $2x+3y+4z+5=0$ and perpendicular to the plane $4x+5y-3z-8=0$ is

• A. $x+7y+13z-96=0$
• B. $x+7y+13z+96=0$
• C. $x+7y-13z-96=0$
• D. $x-7y+13z+96=0$

Answer 1

Answer: (B)

$x+7y+13z+96=0$

The intersection of two planes is
$(x+y+z-6)+\lambda (2x+3y+4z+5)=0$
$\Rightarrow (1+2\lambda )x+(1+3\lambda )y+(1+4\lambda )z+(-6+5\lambda )=\mathrm{0...}\left(i\right)$
Since this plane is perpendicular to the plane
$4x+5y-3z-8=0$
$\therefore (1+2\lambda )4+(1+3\lambda )5+(1+4\lambda )(-3)=0$
$\lambda =\frac{-6}{11}$
On putting the value of $\lambda$ in Eq.(i) we get
$(-\frac{1}{11})x+(-\frac{7}{11})y+(-\frac{13}{11})z+(-\frac{96}{11})=0$
$x+7y+13z+96=0$