Question

Find eqn of plane through line of intersection of planes $x+y+z=12x+3y+4z=5$ which is $\perp lar$ to plane $x-y+z=0$.

Answer 1

$\begin{array}{c}x+y+z-1+\lambda \left(2x+3y+4z-5\right)=0\\ x\left(1+2\lambda \right)+y\left(1+3\lambda \right)+z\left(1+4\lambda \right)=1+5\lambda \\ 1+2\lambda -\left(1+3\lambda \right)+1+4\lambda =0\\ 2\lambda -3\lambda +1+4\lambda =0\\ 3\lambda +1=0\\ \lambda =-\frac{1}{3}\\ \frac{x}{3}+y\left(0\right)+z\left(1-\frac{4}{3}\right)=1-\frac{5}{3}\\ \frac{x}{3}+z\left(-\frac{1}{3}\right)=-\frac{2}{3}\\ x-z=-2\\ x-z+2=0\end{array}$