The equation of the plane that contains the line of intersection of the planes. $x+2 y+3 z-4=0$ and $2…
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
$14 x+7 y-7 z-4=0$
$33 x+45 y+50 z-41=0$
$-33 \mathrm{x}+45-50 \mathrm{z}+41=0$
$5 x+31 y+50 z-41=0$
Solution
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
$\begin{aligned}
& \left(1+\frac{14}{19}\right) \mathrm{x}+\left(2+\frac{7}{19}\right) \mathrm{y}+\left(3-\frac{7}{19}\right) \mathrm{z}+\left(-4+\frac{35}{19}\right)=0 \\
& \therefore 33 \mathrm{x}+45 \mathrm{y}+50 \mathrm{z}-41=0
\end{aligned}$