## Problem Statement

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

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

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

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

$M_{0}(5 ;-4 ; 5)$
## Solution

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

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

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-3 & z-6 \\
1 & -1 & -5 \\
-2 & -3 & -5
\end{array}\right|=0 \\
& (x-1) \cdot\left|\begin{array}{cc}
-1 & -5 \\
-3 & -5
\end{array}\right|-(y-3) \cdot\left|\begin{array}{cc}
1 & -5 \\
-2 & -5
\end{array}\right|+(z-6) \cdot\left|\begin{array}{cc}
1 & -1 \\
-2 & -3
\end{array}\right|=0 \\
& (x-1) \cdot(-10)-(y-3) \cdot(-15)+(z-6) \cdot(-5)=0
\end{aligned}
$$

$-10 x+10+15 y-45-5 z+30=0$

$-10 x+15 y-5 z-5=0$

$2 x-3 y+z+1=0$

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:

$$
d=\frac{|2 \cdot 5-3 \cdot(-4)+5+1|}{\sqrt{(2)^{2}+(-3)^{2}+1^{2}}}=\frac{|10+12+5+1|}{\sqrt{4+9+1}}=\frac{28}{\sqrt{14}}=2 \sqrt{14}
$$

## Problem Kuznetsov Analytic Geometry 8-16