Find the equation of the plane through the line of intersection of the planes

Solution:

The equation of the plane through the intersection of the planes, $x+y+z=1$ and $2 x+3 y+4 z=5$, is

$(x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0$

$\Rightarrow(2 \lambda+1) x+(3 \lambda+1) y+(4 \lambda+1) z-(5 \lambda+1)=0$           ...(1)

The direction ratios, $a_{1}, b_{1}, c_{1}$, of this plane are $(2 \lambda+1),(3 \lambda+1)$, and $(4 \lambda+1)$.

The plane in equation (1) is perpendicular to $x-y+z=0$

Its direction ratios, $a_{2}, b_{2}, c_{2}$, are $1,-1$, and 1 .

Since the planes are perpendicular,

$a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$

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

Substituting $\lambda=-\frac{1}{3}$ in equation $(1)$, we obtain

$\frac{1}{3} x-\frac{1}{3} z+\frac{2}{3}=0$

$\Rightarrow x-z+2=0$

This is the required equation of the plane.

$\Rightarrow 3 \lambda+1=0$

$\Rightarrow \lambda=-\frac{1}{3}$