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.
  
• B.
  
• C.
  
• D. None of these

Answer 1

The correct option is A

Equation of planes passing through intersecting the planes 3x - y - 4z = 0 and x + 3y + 6 = 0 is, (3x - y - 4z) + $\lambda$ (x + 3y + 6) = 0 .....(i)
Given, distance of plane (i) from origin is 1.
$\therefore \frac{6\lambda }{\sqrt{(3+\lambda {)}^2+(3\lambda -1{)}^2+4^2}}$
$or36{\lambda }^2=10{\lambda }^2+26or\lambda =\pm 1$
Put the value of $\lambda$ in (i),
$\therefore$ (3x - y - 4z) $\pm$ (x + 3y + 6) = 0
or 4x + 2y - 4z + 6 = 0 or 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.