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

The equation of the required plane
\(\begin{aligned}
& (x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 \\
& \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1+4 \lambda) z-1-5 \lambda=0
\end{aligned}\)
Let \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) be the d.r.s. of the required plane.
\(\therefore\) From (i), \(a=1+2 \lambda, b=1+3 \lambda, \mathrm{c}=1+4 \lambda\) The required plane is perpendicular to \(x-y+z=0\)
\(\therefore a-b+c=0 \)
\( \Rightarrow 1+2 \lambda-(1+3 \lambda)+1+4 \lambda=0 \)
\( \Rightarrow 1+3 \lambda=0 \)
\( \Rightarrow \lambda=-\frac{1}{3}\)
Substituting \(\lambda=-\frac{1}{3}\) in (i), we get
\(\left(1-\frac{2}{3}\right) x+\left(1-\frac{3}{3}\right) y+\left(1-\frac{4}{3}\right) z-1~+\) \(\frac{5}{3}=0 \)
\( \Rightarrow x-z+2=0\)
Its vector equation is
\(\overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{k}})+2=0\)