Question

The equation of the plane 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, is

• 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. None of these

Answer 1

The correct option is A x -2y -2z-3 =0, 2x+y -2z+3 =0

Equation of planes passing through intersection of the planes 3x-y-4z=0 and x+3y+6=0 is, $(3x-y-4z)+\lambda (x+3y+6)=0.....\left(i\right)$
Given, distance of plane (i) from origin is 1.
$\therefore \frac{6\lambda }{\sqrt{(3+\lambda {)}^2+(3\lambda -1{)}^2+4^2}}=1$
or $36{\lambda }^2=10{\lambda }^2+26$ or $\lambda =\pm 1$
Put the value of $\lambda$ in (i),
$\therefore (3x-y-4z)\pm (x+3y+6)=0$
or 4x+2y -4z+6=0 i.e 2x+y-2z+3 =0
and 2x-4y-4z-6=0
Thus the required planes are x-2y-2z-3=0 and 2x+y-2z+3=0.