Question

Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.

Answer 1

$$\begin{gathered} \text{The equation of any plane passing through (1, -3, -2) is} \\ a\left(x-1\right)+b\left(y+3\right)+c\left(z+2\right)=0...\left(1\right) \\ \text{It is given that (1) is perpendicular to the planes}x+2y+2z=5\text{and 3}x+3y+2z=8.\text{Then,} \\ a+2b+2c=0...\left(2\right) \\ 3a+3b+2c=0...\left(3\right) \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x-1& y+3& z+2\\ 1& 2& 2\\ 3& 3& 2\end{array}\right|=0 \\ \Rightarrow -2\left(x-1\right)+4\left(y+3\right)-3\left(z+2\right)=0 \\ \Rightarrow -2x+2+4y+12-3z-6=0 \\ \Rightarrow 2x-4y+3z-8=0 \end{gathered}$$