## problem statement

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

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

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

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

$M_{0}(1 ;-6 ;-5)$
## Solution

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

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

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+1 & y-2 & z+3 \\
5 & -3 & 3 \\
3 & -1 & 1
\end{array}\right|=0 \\
& (x+1) \cdot\left|\begin{array}{cc}
-3 & 3 \\
-1 & 1
\end{array}\right|-(y-2) \cdot\left|\begin{array}{cc}
5 & 3 \\
3 & 1
\end{array}\right|+(z+3) \cdot\left|\begin{array}{cc}
5 & -3 \\
3 & -1
\end{array}\right|=0 \\
& (x+1) \cdot 0-(y-2) \cdot(-4)+(z+3) \cdot 4=0 \\
& 4 y-8+4 z+12=0 \\
& 4 y+4 z+4=0 \\
& y+z+1=0
\end{aligned}
$$

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:

$$
d=\frac{|1 \cdot(-6)+1 \cdot(-5)+1|}{\sqrt{0^{2}+1^{2}+1^{2}}}=\frac{|-6-5+1|}{\sqrt{0+1+1}}=\frac{10}{\sqrt{2}}=5 \sqrt{2}
$$

## Problem Kuznetsov Analytic Geometry 8-2