## Problem Statement

Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.

$M_{1}(0 ;-1 ;-1)$
$M_{2}(-2 ; 3 ; 5)$
$M_{3}(1 ;-5 ;-9)$
$M_{0}(-4 ;-13 ; 6)$
## Solution

Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:

$$
\left|\begin{array}{ccc}
x-0 & y-(-1) & z-(-1) \\
-2-0 & 3-(-1) & 5-(-1) \\
1-0 & -5-(-1) & -9-(-1)
\end{array}\right|=0
$$

Perform transformations:

$\left|\begin{array}{ccc}x & y+1 & z+1 \\ -2 & 4 & 6 \\ 1 & -4 & -8\end{array}\right|=0$

$x \cdot\left|\begin{array}{cc}4 & 6 \\ -4 & -8\end{array}\right|-(y+1) \cdot\left|\begin{array}{cc}-2 & 6 \\ 1 & -8\end{array}\right|+(z+1) \cdot\left|\begin{array}{cc}-2 & 4 \\ 1 & -4\end{array}\right|=0$

$x \cdot(-8)-(y+1) \cdot 10+(z+1) \cdot 4=0$

$-8 x-10 y-10+4 z+4=0$

$-8 x-10 y+4 z-6=0$

$-4 x-5 y+2 z-3=0$

The distance $d$ from a point $M_{0}\left(x_{0} ; y_{0} ; z_{0}\right)$ to the plane $A x+B y+C z+D=0$:

$d=\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$

Find:

$$
d=\frac{|-4 \cdot(-4)-5 \cdot(-13)+2 \cdot 6-3|}{\sqrt{(-4)^{2}+(-5)^{2}+2^{2}}}=\frac{|16+65+12-3|}{\sqrt{16+25+4}}=\frac{90}{\sqrt{45}}=2 \sqrt{45}
$$

## Problem Kuznetsov Analytic Geometry 8-21