## Problem Statement

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

$M_{1}(5 ; 2 ; 0)$

$M_{2}(2 ; 5 ; 0)$

$M_{3}(1 ; 2 ; 4)$

$M_{0}(-3 ;-6 ;-8)$
## Solution

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

$$
\left|\begin{array}{lll}
x-5 & y-2 & z-0 \\
2-5 & 5-2 & 0-0 \\
1-5 & 2-2 & 4-0
\end{array}\right|=0
$$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-5 & y-2 & z \\
-3 & 3 & 0 \\
-4 & 0 & 4
\end{array}\right|=0 \\
& (x-5) \cdot\left|\begin{array}{cc}
3 & 0 \\
0 & 4
\end{array}\right|-(y-2) \cdot\left|\begin{array}{cc}
-3 & 0 \\
-4 & 4
\end{array}\right|+z \cdot\left|\begin{array}{cc}
-3 & 3 \\
-4 & 0
\end{array}\right|=0 \\
& (x-5) \cdot 12-(y-2) \cdot(-12)+z \cdot 12=0 \\
& 12 x-60+12 y-24+12 z=0 \\
& 12 x+12 y+12 z-84=0 \\
& x+y+z-7=0
\end{aligned}
$$

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:

$$
d=\frac{|-3-6-8-7|}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{|-24|}{\sqrt{1+1+1}}=\frac{24}{\sqrt{3}}=8 \sqrt{3}
$$

## Problem Kuznetsov Analytic Geometry 8-22