Question

The equation of the plane passing through the line of intersection of the planes $x+y+z=1,2x+3y+4z=7$, and perpendicular to the plane $x-5y+3x=5$ is given by

• A. $x+2y+3z-6=0$
• B. $x+2y+3z+6=0$
• C. $3x+4y+5z-8=0$
• D. $3x+4y+5z+8=0$

Answer 1

The correct option is A $x+2y+3z-6=0$

Let the equation of plane be $P_1+\lambda P_2=0$

$(x+y+z-1)+\lambda (2x+3y+4z-7)=0$

Or, $(2\lambda +1)x+(3\lambda +1)y+(4\lambda +1)z=7\lambda +1$

This plane is perpendicular to the plane $x-5y+3x=5$

$\therefore 1(2\lambda +1)+(-5)(3\lambda +1)+3(4\lambda +1)=0$

Or, $\lambda =-1$

$\therefore$ plane is $-x-2y-3z=-6$ or $x+2y+3z-6=0$