Question

Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.

Answer 1

$$\begin{gathered} \text{The equation of any plane passing through (-1, -1, 2) is} \\ a\left(x+1\right)+b\left(y+1\right)+c\left(z-2\right)=0...\left(1\right) \\ \text{It is given that (1) is perpendicular to each of the planes}3x+2y-3z=1\text{and}5x-4y+z=5.\text{Then,} \\ 3a+2b-3c=0...\left(2\right) \\ 5a-4b+c=0...\left(3\right) \\ \text{Solving (1), (2) and (3), we get} \\ \left|\begin{array}{ccc}x+1& y+1& z-2\\ 3& 2& -3\\ 5& -4& 1\end{array}\right|=0 \\ \Rightarrow -10\left(x+1\right)-18\left(y+1\right)-22\left(z-2\right)=0 \\ \Rightarrow 5\left(x+1\right)+9\left(y+1\right)+11\left(z-2\right)=0 \\ \Rightarrow 5x+5+9y+9+11z-22=0 \\ \Rightarrow 5x+9y+11z-8=0 \end{gathered}$$