## Problem Statement

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

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

$M_{2}(-2 ; 0 ; 3)$

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

$M_{0}(-2 ; 4 ; 2)$
## Solution

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

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

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y+1 & z-1 \\
-3 & 1 & 2 \\
1 & 2 & -2
\end{array}\right|=0 \\
& (x-1) \cdot\left|\begin{array}{cc}
1 & 2 \\
2 & -2
\end{array}\right|-(y+1) \cdot\left|\begin{array}{cc}
-3 & 2 \\
1 & -2
\end{array}\right|+(z-1) \cdot\left|\begin{array}{cc}
-3 & 1 \\
1 & 2
\end{array}\right|=0 \\
& (x-1) \cdot(-6)-(y+1) \cdot 4+(z-1) \cdot(-7)=0 \\
& -6 x+6-4 y-4-7 z+7=0 \\
& -6 x-4 y-7 z+9=0
\end{aligned}
$$

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:

$$
d=\frac{|-6 \cdot(-2)-4 \cdot 4-7 \cdot 2+9|}{\sqrt{(-6)^{2}+(-4)^{2}+(-7)^{2}}}=\frac{|12-16-14+9|}{\sqrt{36+16+49}}=\frac{9}{\sqrt{101}}
$$

## Problem Kuznetsov Analytic Geometry 8-4