Question:
The vector equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2 x+3 y+4 z=5$ which is perpendicular to the plane $x-y+z=0$ is :
Correct Option: , 3
Solution:
Let the plane be
$(x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0$
$\Rightarrow(2 \lambda+1) \mathrm{x}+(3 \lambda+1) \mathrm{y}+(4 \lambda+1) \mathrm{z}-(5 \lambda+1)=0$
$\perp$ to the plane $x-y+z=0$
$\Rightarrow \lambda=-\frac{1}{3}$
$\Rightarrow$ the required plane is $\mathrm{x}-\mathrm{z}+2=0$