Question

The equation of the planes passing through the line of intersection of the planes $3x-y-4z=0$ and $x+3y+6=0$, whose distance from the origin is $1$, are

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

Answer 1

Answer: (A)

$x-2y-2z-3=0,2x+y-2z+3=0$

Plane containing line of intersection of $3x-y-4z=0$ & $x+3y+6=0$

$3x-y-4z+\lambda (x+3y+6)=0(3+\lambda )x+(3\lambda -1)y+(-4)z+6\lambda =0$

Distance from origin is $\perp$ unit.

$\therefore \frac{(3+\lambda )\left(0\right)+(3\lambda -1)\left(0\right)+(-4)\left(0\right)+6\lambda }{\sqrt{{(3+\lambda )}^2+{(3\lambda -1)}^2+{(-4)}^2}}=1\Rightarrow {\left(6\lambda \right)}^2={(3+\lambda )}^2+{(3\lambda -1)}^2+4^2\Rightarrow 36{\lambda }^2=9+{\lambda }^2+6\lambda +9{\lambda }^2+1-6\lambda +16\Rightarrow \lambda =\pm 1$

$\therefore$ Planes possible,

$4x+2y-4z+6=0\Rightarrow 2x+y-2z+3=02x-4y-4z-6=0\Rightarrow x-2y-2z-3=0$