Question

Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.

Answer 1

$$\begin{gathered} \text{The equation of any plane passing through (1, -1, 2) is} \\ a\left(x-1\right)+b\left(y+1\right)+c\left(z-2\right)=0...\left(1\right) \\ \text{It is given that (1) is perpendicular to the plane 2}x+3y-2z=5.\mathrm{So}, \\ 2a+3b-2c=0...\left(2\right) \\ \text{It is given that (1) is perpendicular to the plane}x+2y-3z=8.\text{So,} \\ a+2b-3c=0...\left(3\right) \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x-1& y+1& z-2\\ 2& 3& -2\\ 1& 2& -3\end{array}\right|=0 \\ \Rightarrow -5\left(x-1\right)+4\left(y+1\right)+1\left(z-2\right)=0 \\ \Rightarrow 5x-4y-z=7 \end{gathered}$$