## Problem Statement

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

$M_{1}(-1 ; 2 ; 4)$

$M_{2}(-1 ;-2 ;-4)$

$M_{3}(3 ; 0 ;-1)$

$M_{0}(-2 ; 3 ; 5)$
## Solution

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

$\left|\begin{array}{ccc}x-(-1) & y-2 & z-4 \\ -1-(-1) & -2-2 & -4-4 \\ 3-(-1) & 0-2 & -1-4\end{array}\right|=0$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+1 & y-2 & z-4 \\
0 & -4 & -8 \\
4 & -2 & -5
\end{array}\right|=0 \\
& (x+1) \cdot\left|\begin{array}{cc}
-4 & -8 \\
-2 & -5
\end{array}\right|-(y-2) \cdot\left|\begin{array}{cc}
0 & -8 \\
4 & -5
\end{array\right|+(z-4) \cdot\left|\begin{array}{cc}
0 & -4 \\
4 & -2
\end{array}\right|=0 \\
& (x+1) \cdot 4-(y-2) \cdot 32+(z-4) \cdot 16=0 \\
& 4 x+4-32 y+64+16 z-64=0 \\
& 4 x-32 y+16 z+4=0 \\
& x-8 y+4 z+1=0
\end{aligned}
$$

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:

$$
d=\frac{|-2-8 \cdot 3+4 \cdot 5+1|}{\sqrt{1^{2}+(-8)^{2}+4^{2}}}=\frac{|-2-24+20+1|}{\sqrt{1+64+16}}=\frac{5}{\sqrt{81}}=\frac{5}{9}
$$

## Problem Kuznetsov Analytic Geometry 8-9