Question

The equation of the plane which contains the line of 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 is :

• A. 51x +15y – 50z + 173 = 0
• B. 51x + 15y – 50z = 0
• C. 51x +15y – 50z + 10 = 0
• D. 51x +15y – 50z + 30 = 0
  

Answer 1

The correct option is A 51x +15y – 50z + 173 = 0

Equation of plane through in the line of intersection of the planes
x + 2y + 3z - 4 = 0, 2x + y - z + 5 = 0 is
$(x+2y+3z-4)+\lambda (2x+y-z+5)=0$
or $(1+2\lambda )x+2(2+\lambda )y+(3-\lambda )z+(5\lambda -4)=0....\left(1\right)$
Plane (1) is perpendicular to plane 5x + 3y + 6z + 8 = 0
$\therefore 5(1+2\lambda )+3(2+\lambda )z+(5\lambda -4)=0$
$\therefore \lambda =-\frac{29}{7}$
Substituting this value of λ in (1), we have the equation of required plane is 51x +15y - 50z + 173 = 0.