Question

Find the equation of the plane passing through the points whose coordinates are (−1, 1, 1) and (1, −1, 1) and perpendicular to the plane x + 2y + 2z = 5.

Answer 1

$$\begin{gathered} \text{The equation of any plane passing through (-1, 1, 1) is} \\ a\left(x+1\right)+b\left(y-1\right)+c\left(z-1\right)=0...\left(1\right) \\ \text{It is given that (1) is passing through (1, -1, 1). So,} \\ a\left(1+1\right)+b\left(-1-1\right)+c\left(1-1\right)=0 \\ \Rightarrow 2a-2b+0c=0...\left(2\right) \\ \text{It is given that (1) is perpendicular to the plane}x+2y+2z=5.\text{So,} \\ a+2b+2c=0...\left(3\right) \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x+1& y-1& z-1\\ 2& -2& 0\\ 1& 2& 2\end{array}\right|=0 \\ \Rightarrow -4\left(x+1\right)-4\left(y-1\right)+6\left(z-1\right)=0 \\ \Rightarrow 2\left(x+1\right)+2\left(y-1\right)-3\left(z-1\right)=0 \\ \Rightarrow 2x+2y-3z+3=0 \end{gathered}$$