## Problem Statement

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

$M_{1}(-2 ; 0 ;-4)$

$M_{2}(-1 ; 7 ; 1)$

$M_{3}(4 ;-8 ;-4)$

$M_{0}(-6 ; 5 ; 5)$
## Solution

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

$$
\left|\begin{array}{ccc}
x-(-2) & y-0 & z-(-4) \\
-1-(-2) & 7-0 & 1-(-4) \\
4-(-2) & -8-0 & -4-(-4)
\end{array}\right|=0
$$

Perform the transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+2 & y & z+4 \\
1 & 7 & 5 \\
6 & -8 & 0
\end{array}\right|=0 \\
& (x+2) \cdot\left|\begin{array}{cc}
7 & 5 \\
-8 & 0
\end{array}\right|-y \cdot\left|\begin{array}{cc}
1 & 5 \\
6 & 0
\end{array}\right|+(z+4) \cdot\left|\begin{array}{cc}
1 & 7 \\
6 & -8
\end{array}\right|=0 \\
& (x+2) \cdot 40-y \cdot(-30)+(z+4) \cdot(-50)=0
\end{aligned}
$$

$40 x+80+30 y-50 z-200=0$

$40 x+30 y-50 z-120=0$

$4 x+3 y-5 z-12=0$

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:

$$
d=\frac{\mid 4 \cdot(-6)+3 \cdot 5-5 \cdot 5-12) \mid}{\sqrt{4^{2}+3^{2}+(-5)^{2}}}=\frac{|-24+15-25-12|}{\sqrt{16+9+25}}=\frac{46}{\sqrt{50}}=\frac{46}{5 \sqrt{2}}=\frac{23 \sqrt{2}}{5}
$$

## Problem Kuznetsov Analytic Geometry $8-24$