Question

Find the equation of the plane passing through the intersection of the planes 3x-y+2z-4=0 and x+y+z-2=0 and the point (2,2,1)

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{planes}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{planes} \\ ax+by+cz+d=0\mathrm{and}{a}_{1}x+b_{1}y+c_{1}z+d_1=0\mathrm{is}\mathrm{given}\mathrm{by}, \\ ax+by+cz+d+\lambda \left(a_{1}x+b_{1}y+c_{1}z+d_1\right)=0 \\ \mathrm{So},\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{plane}\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{planes} \\ 3x-y+2z-4=0andx+y+z-2=0is \\ \left(3x-y+2z-4\right)+\lambda \left(x+y+z-2\right)=0 \\ \Rightarrow \left(3+\lambda \right)x+\left(\lambda -1\right)y+\left(2+\lambda \right)z-4-2\lambda =0.........\left(1\right) \\ \mathrm{Since}\mathrm{the}\mathrm{plane}\left(1\right)\mathrm{passes}\mathrm{through}\left(2,2,1\right),\mathrm{then}\mathrm{it}\mathrm{must}\mathrm{satisfy}\mathrm{it}. \\ \mathrm{Put}x=2;y=2;z=1\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \left(3+\lambda \right)2+\left(\lambda -1\right)2+\left(2+\lambda \right)1-4-2\lambda =0 \\ \Rightarrow 6+2\lambda +2\lambda -2+2+\lambda -4-2\lambda =0 \\ \Rightarrow 3\lambda +2=0 \\ \Rightarrow \lambda =-\frac{2}{3} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{\lambda }\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \left(3-\frac{2}{3}\right)x+\left(-\frac{2}{3}-1\right)y+\left(2-\frac{2}{3}\right)z-4+\frac{4}{3}=0 \\ \Rightarrow \frac{7x}{3}-\frac{5y}{3}+\frac{4z}{3}-\frac{8}{3}=0 \\ \Rightarrow \boxed{7x-5y+4z-8=0} \end{gathered}$$