## Problem Statement

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

$M_{1}(-1 ; -5 ; 2)$
$M_{2}(-6 ; 0 ; -3)$
$M_{3}(3 ; 6 ; -3)$
$M_{0}(10 ; -8 ; -7)$
## Solution

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

$$
\left|\begin{array}{ccc}
x-(-1) & y-(-5) & z-2 \\
-6-(-1) & 0-(-5) & -3-2 \\
3-(-1) & 6-(-5) & -3-2
\end{array}\right|=0
$$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+1 & y+5 & z-2 \\
-5 & 5 & -5 \\
4 & 11 & -5
\end{array}\right|=0 \\
& (x+1) \cdot\left|\begin{array}{cc}
5 & -5 \\
11 & -5
\end{array}\right|-(y+5) \cdot\left|\begin{array}{cc}
-5 & -5 \\
4 & -5
\end{array}\right|+(z-2) \cdot\left|\begin{array}{cc}
-5 & 5 \\
4 & 11
\end{array}\right|=0 \\
& (x+1) \cdot 30-(y+5) \cdot 45+(z-2) \cdot(-75)=0 \\
& 30 x+30-45 y-225-75 z+150=0 \\
& 30 x-45 y-75 z-45=0 \\
& 2 x-3 y-5 z-3=0
\end{aligned}
$$

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:

$$
d=\frac{|2 \cdot 10-3 \cdot(-8)-5 \cdot(-7)-3|}{\sqrt{2^{2}+(-3)^{2}+(-5)^{2}}}=\frac{\mid 20+24+35-3) \mid}{\sqrt{4+9+25}}=\frac{76}{\sqrt{38}}=2 \sqrt{38}
$$

## Problem Kuznetsov Analytic Geometry $8-20$