Question

Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.

Answer 1

$$\begin{gathered} \text{The equation of any plane passing through (2, 2, 1) is} \\ a\left(x-2\right)+b\left(y-2\right)+c\left(z-1\right)=0...\left(1\right) \\ \text{It is given that (1) is passing through (9, 3, 6). So,} \\ a\left(9-2\right)+b\left(3-2\right)+c\left(6-1\right)=0 \\ \Rightarrow 7a+b+5c=0...\left(2\right) \\ \text{It is given that (1) is perpendicular to the plane 2}x+6y+6z=1.\text{So,} \\ 2a+6b+6c=0 \\ \Rightarrow a+3b+3c=0...\left(3\right) \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x-2& y-2& z-1\\ 7& 1& 5\\ 1& 3& 3\end{array}\right|=0 \\ \Rightarrow -12\left(x-2\right)-16\left(y-2\right)+20\left(z-1\right)=0 \\ \Rightarrow 3\left(x-2\right)+4\left(y-2\right)-5\left(z-1\right)=0 \\ \Rightarrow 3x+4y-5z=9 \end{gathered}$$