Question

The equation of the plane passing through the line of intersection of the planes $x+y+z=$ $6,2x+3y+4z+5=0$ and through the point $(1,1,1)$

• A. $2x+3y+4z=9$
• B. $3x+4y-8z+1=0$
• C. $20x+23y+26z=69$
• D. $4x-y+3z-6=0$

Answer 1

The correct option is C $20x+23y+26z=69$

The equation of plane passing through intersection of plane $x+y+z=6$ and $2x+3y+4z+5=0$ is given by,
$(x+y+z-6)+\lambda (2x+3y+4z+5)=0$
Also given this plane is passing through $(1,1,1)$.
$\Rightarrow (1+1+1-6)+\lambda (2+3+4+5)=0\Rightarrow \lambda =\frac{3}{14}$
Hence required plane is $20x+23y+26z=69$.