## Problem Statement

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

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

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

$M_{3}(3 ; 2 ; 1)$

$M_{0}(-5 ; 3 ; 7)$
## Solution

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

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

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y+1 & z-2 \\
-1 & 3 & -3 \\
1 & 3 & -1
\end{array}\right|=0 \\
& (x-2) \cdot\left|\begin{array}{cc}
3 & -3 \\
3 & -1
\end{array}\right|-(y+1) \cdot\left|\begin{array}{cc}
-1 & -3 \\
1 & -1
\end{array}\right|+(z-2) \cdot\left|\begin{array}{cc}
-1 & 3 \\
1 & 3
\end{array}\right|=0 \\
& (x-2) \cdot 6-(y+1) \cdot 4+(z-2) \cdot(-6)=0 \\
& 6 x-12-4 y-4-6 z+12=0 \\
& 6 x-4 y-6 z-4=0 \\
& 3 x-2 y-3 z-2=0
\end{aligned}
$$

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

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

## Find:

$$
d=\frac{|3 \cdot(-5)-2 \cdot 3-3 \cdot 7-2|}{\sqrt{3^{2}+(-2)^{2}+(-3)^{2}}}=\frac{|-15-6-21-2|}{\sqrt{9+4+9}}=\frac{44}{\sqrt{22}}=2 \sqrt{22}
$$

## Problem Kuznetsov Analytic Geometry 8-27