Question

The equation of the plane which contains the line of inetrsection 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. $51x-15y-50z-10=0$

Answer 1

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

Plane1:$x+2y+3z-4=0$

Plane2:$2x+y-z+5=0$

Equation of plane through line of intersection of plane1 & plane2 is,

$(x+2y+3z-4)+\lambda (2x+y-z+5)=0(1+2\lambda )x+(2+\lambda )y+(3-\lambda )z+5\lambda -4=0$

$\therefore$ Directions of normal to plane generated $=(1+2\lambda ,2+\lambda ,3-\lambda )$

The new plane has to be $\perp$ to $5x+3y+6z+8=0$

$\therefore (1+2\lambda )5+(2+\lambda )3+(3-\lambda )6=0\Rightarrow \lambda =\frac{-29}{7}$

The equation of new plane,

$\frac{-51}{7}x-\frac{15}{7}y+\frac{50}{7}z-\frac{173}{7}=051x+15y-50z+173=0$