Question

Find the equation of the plane through the intersection of the planes $x+3y+6=0$ and $3x-y-4z=0,$ whose perpendicular distance from the origin is unity.

• A. $x-2y-2z-3=0.$
• B. $2x+y-2z+3=0$
• C. $2x+2y-z+3=0$
• D. both A and B

Answer 1

The correct option is D both A and B

Any plane through the intersection of given planes is
$(x+3y+6)+\lambda (3x-y-4z)=0$
$(1+3\lambda )x+(3-\lambda )y-4\lambda z+6=0.$ ...(1)
Its perpendicular distance from $(0,0,0)$ is $1$.
$\therefore \frac{6}{{[{(1+3\lambda )}^2+{(3-\lambda )}^2+16{\lambda }^2]}^{1/2}}=1.$
$1+6\lambda +9{\lambda }^2+9+{\lambda }^2-6\lambda +16{\lambda }^2=36$
$\therefore 26{\lambda }^2=26$ or $\lambda =\pm 1.$
Putting the value of $\lambda$ in ( 1 ), the required equations are
$2x+y-2z+3=0$ and $x-2y-2z-3=0.$