Q.Find the equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to the plane $x-y+z=0$.

Write the family of planes through the line of intersection, then use the perpendicularity condition (normal $\cdot$ normal $=0$) to fix the parameter.

Planes: $x+y+z=1$ and $2x+3y+4z=5$.

Family of planes through their line of intersection:

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

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

Its normal vector is $\vec n_1=(1+2\lambda,\ 1+3\lambda,\ 1+4\lambda)$.

This plane must be perpendicular to $x-y+z=0$, whose normal is $\vec n_2=(1,-1,1)$. Perpendicularity of planes means $\vec n_1\cdot\vec n_2=0$:

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

$1+2\lambda-1-3\lambda+1+4\lambda=0$

$1+3\lambda=0 \ \Rightarrow\ \lambda=-\frac13$

Substitute back into (1):

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