Question

Find the equation of the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9.

Answer 1

$$\begin{gathered} \text{The equation of any plane passing through (1, -1, 2) is} \\ a\left(x-1\right)+b\left(y+1\right)+c\left(z-2\right)=0...\left(1\right) \\ \text{It is given that (1) is passing through (2, -2, 2). So,} \\ a\left(2-1\right)+b\left(-2+1\right)+c\left(2-2\right)=0 \\ \Rightarrow a-b+0c=0...\left(2\right) \\ \text{It is given that (1) is perpendicular to the plane 6}x-2y+2z=9.\text{So,} \\ 6a-2b+2c=0 \\ \Rightarrow 3a-b+c=0...\left(3\right) \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x-1& y+1& z-2\\ 1& -1& 0\\ 3& -1& 1\end{array}\right|=0 \\ \Rightarrow -1\left(x-1\right)-1\left(y+1\right)+2\left(z-2\right)=0 \\ \Rightarrow -x+1-y-1+2z-4=0 \\ \Rightarrow x+y-2z+4=0 \end{gathered}$$