The equation of the plane through the intersection of the planes $\mathrm{x}+\mathrm{y}+\mathrm{z}=1$ and $2…
The equation of the plane through the intersection of the planes $\mathrm{x}+\mathrm{y}+\mathrm{z}=1$ and $2 \mathrm{x}+3 \mathrm{y}-\mathrm{x}+4=0$ and parallel to $\mathrm{X}$-axis is
$y+3 z+6=0$
$3 y-z+6=0$
$y-3 z+6=0$
$3 y-2 z+6=0$
Solution
$\begin{aligned} & (x+y+z-1)+\lambda(2 x+3 y-z+4)=0 \\ & \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1-\lambda) z+(4 \lambda-1)=0\end{aligned}$
To be parallel to the $x$-axis $1+2 \lambda=0 \Rightarrow \lambda=\frac{-1}{2}$
$\Rightarrow-\frac{1}{2} y+\frac{3}{2} z-3=0 \Rightarrow y-3 z+6=0$