The equation of the plane that contains the line of intersection of the planes. \(x+2 y+3 z-4=0\) and \(2 x+y+5=0\) and is perpendicular to the plane \(5 x+3 y-6 z+8=0\) is

The equation of the required plane is \((x+2 y+3 z-4)+\lambda\)
\(
\begin{aligned}
& (2 x+y-z+5)=0 \text { i.e. } \\
& (1+2 \lambda) x+(2+\lambda) y+(3-\lambda) z+(-4+5 \lambda)=0
\end{aligned}
\)
Since (1) is perpendicular to the plane \(5 x+3 y-6 z+8=0\), we write
\(
\begin{aligned}
& (1+2 \lambda)(5)+(2+\lambda)(3)+(3-\lambda)(-6)=0 \\
& \therefore 5+10 \lambda+6+3 \lambda-18+6 \lambda=0 \\
& 19 \lambda=7 \\
& \Rightarrow \lambda=\frac{7}{19}
\end{aligned}
\)
Substituting value of \(\lambda\) in eq. (1), we get
\(\left(1+\frac{14}{19}\right) \mathrm{x}+\left(2+\frac{7}{19}\right) \mathrm{y}+\left(3-\frac{7}{19}\right) \mathrm{z}~+\) \(\left(-4+\frac{35}{19}\right)=0\)
\(\therefore 33 \mathrm{x}+45 \mathrm{y}+50 \mathrm{z}-41=0\)