Question

Let the equation of the plane through the intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ and perpendicular to the plane $5x+3y+6z+8=0$ be $kx+15y+mz+173=0$. Find $k+m$

Answer 1

Any plane through the intersection of the given planes is
$(x+2y+3z-4)+\lambda (2x+y-z+5)=0$
or $x(1+2\lambda )+y(2+\lambda )+z(3-\lambda )-(4-5\lambda )=0$ ...( 1 )
Since the plane ( 1 ) is perpendicular to the plane
$5x+3y+6z+8=0$
$\therefore 5(1+2\lambda )+3(2+\lambda )+6(3-\lambda )=0$
$29+7\lambda =0;\lambda =-29/7.$
Substituting the value of $\lambda$ in ( 1 ), we get the equation of the required plane as
$-51x-15y+50z-173=0$
$\Rightarrow 51x+15y-50z+173=0$
$\Rightarrow m+k=1$