Question

What is the equation of the plane passing through the line of intersection of the planes $x-y+3z=4$ and $2x+y+3z=5$ and parallel to the plane $x+y+z=1$?

• A. $x+y+z=2$
• B. $x+y+z+2=0$
• C. $2x=y+z$
• D. no plane exists

Answer 1

The correct option is A $x+y+z=2$

The equation of the plane passing through the line of intersection of the planes $x-y+3z=4$ and $2x+y+3z=5$ is

$x-y+3z-4+k(2x+y+3z-5)=0$

$\Rightarrow (1+2k)x+(-1+k)y+(3+3k)=4+5k$

Since, above plane is parallel with $x+y+z=1$

Therefore, $1+2k=-1+k=3+3k$

$\Rightarrow k=-2$

Therefore, equation of desired plane is $(1+2(-2\left)\right)x+(-1+(-2\left)\right)y+(3+3(-2\left)\right)=4+5(-2)$

$\Rightarrow x+y+z=2$
Ans: A