Question

Find the equation of the plane determined by the points A (3, −1, 2), B (5, 2, 4) and C (−1, −1, 6) and, hence, the distance between the plane and the point P (6, 5, 9).

Answer 1

The equation of the plane passing through A (3, −1, 2), B (5, 2, 4) and C (−1, −1, 6) ) is
$$\begin{gathered} \left|\begin{array}{ccc}x-3& y+1& z-2\\ 5-3& 2+1& 4-2\\ -1-3& -1+1& 6-2\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}x-3& y+1& z-2\\ 2& 3& 2\\ -4& 0& 4\end{array}\right|=0 \\ \Rightarrow 12\left(x-3\right)-16\left(y+1\right)+12\left(z-2\right)=0 \\ \Rightarrow 3\left(x-3\right)-4\left(y+1\right)+3\left(z-2\right)=0 \\ \Rightarrow 3x-4y+3z-19=0...\left(1\right) \\ \text{We know that the distance of the point}\left(x_1,y_1,z_1\right)\text{from the plane}ax+by+cz+d=0\text{is given by} \\ \frac{\left|a{x}_1+b{y}_1+c{z}_1+d\right|}{\sqrt{a^2+b^2+c^2}} \\ \text{So, the distance of plane (1) from the point}\mathit{\text{P}}\text{(6, 5, 9) is} \\ \frac{\left|3\left(6\right)-4\left(5\right)+3\left(9\right)-19\right|}{\sqrt{9+16+9}} \\ =\frac{\left|6\right|}{\sqrt{34}} \\ =\frac{6}{\sqrt{34}}\text{units} \end{gathered}$$

Disclaimer: The answer given for the second part of this problem in the text book is incorrect.