## Problem Statement

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

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

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

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

$M_{0}(-5 ;-9 ; 1)$
## Solution

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

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

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y & z-2 \\
0 & 2 & -3 \\
1 & -2 & -1
\end{array}\right|=0 \\
& (x-1) \cdot\left|\begin{array}{cc}
2 & -3 \\
-2 & -1
\end{array}\right|-y \cdot\left|\begin{array}{cc}
0 & -3 \\
1 & -1
\end{array}\right|+(z-2) \cdot\left|\begin{array}{cc}
0 & 2 \\
1 & -2
\end{array}\right|=0 \\
& (x-1) \cdot(-8)-y \cdot 3+(z-2) \cdot(-2)=0 \\
& -8 x+8-3 y-2 z+4=0 \\
& -8 x-3 y-2 z+12=0
\end{aligned}
$$

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:

$$
d=\frac{|-8 \cdot(-5)-3 \cdot(-9)-2 \cdot 1+12|}{\sqrt{(-8)^{2}+(-3)^{2}+(-2)^{2}}}=\frac{|40+27-2+12|}{\sqrt{64+9+4}}=\frac{77}{\sqrt{77}}=\sqrt{77}
$$

## Problem Kuznetsov Analytic Geometry 8-6