Question

The plane $x+3y+13=0$ passes through the line of intersection of the planes $2x-8y+4z=p$ and $3x-5y+4z+10=0$. If the plane is perpendicular to the plane $3x-y-2z-4=0$, then the value of $p$ is

• A. $2$
• B. $5$
• C. $9$
• D. $3$
• E. $-1$

Answer 1

The correct option is D $3$

Equation of plane passes through line of intersection of the plane $2x-8y+4z=p$ and $3x-5y+4z+10=0$ is
$(2x-8y+4z-p)+\lambda (3x-5y+4z+10)=0$
$\Rightarrow (2+3\lambda )x+(-8-5\lambda )y+(4+4\lambda )z-p+10\lambda =0$ ...(i)
Given equation of plane is
$x+3y+13=0$ ....(ii)
Equations (i) and (ii) represent same plane.
$\therefore 4+4\lambda =0\Rightarrow \lambda =-1$
Putting $\lambda =-1$ in equation (i), we get
$-x-3y-p-10=0$
$\Rightarrow x+3y+p+10=0$ ...(iii)
Comparing the coefficient of $x,y$ and $z$ constants term for equations (i) and (ii), we get
$p+10=13$
$\Rightarrow p=3$