## Problem Statement

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

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

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

$M_{3}(-2 ;-1 ; 6)$

$M_{0}(3 ;-2 ;-9)$
## Solution

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

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

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-2 & z+3 \\
0 & -2 & 4 \\
-3 & -3 & 9
\end{array}\right|=0 \\
& (x-1) \cdot\left|\begin{array}{ll}
-2 & 4 \\
-3 & 9
\end{array}\right|-(y-2) \cdot\left|\begin{array}{cc}
0 & 4 \\
-3 & 9
\end{array}\right|+(z+3) \cdot\left|\begin{array}{cc}
0 & -2 \\
-3 & -3
\end{array}\right|=0 \\
& (x-1) \cdot(-6)-(y-2) \cdot 12+(z+3) \cdot(-6)=0 \\
& -6 x+6-12 y+24-6 z-18=0 \\
& -6 x-12 y-6 z+12=0 \\
& x+2 y+z-2=0
\end{aligned}
$$

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:

$$
d=\frac{|3+2 \cdot(-2)-9-2|}{\sqrt{1^{2}+2^{2}+1^{2}}}=\frac{|3-4-9-2|}{\sqrt{1+4+1}}=\frac{12}{\sqrt{6}}=2 \sqrt{6}
$$

## Problem Kuznetsov Analytic Geometry 8-7