## Problem Statement

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

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

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

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

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

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

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

Perform transformations:

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

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:

$$
d=\frac{|5 \cdot 2-2 \cdot(-1)-3 \cdot 4-1|}{\sqrt{5^{2}+(-2)^{2}+(-3)^{2}}}=\frac{|10+2-12-1|}{\sqrt{25+4+9}}=\frac{1}{\sqrt{38}}
$$

## Problem Kuznetsov Analytic Geometry 8-5