## problem statement

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

$M_{1}(1 ; 5 ;-7)$

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

$M_{3}(-2 ; 7 ; 3)$

$M_{0}(1 ;-1 ; 2)$
## Solution

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

$\left|\begin{array}{ccc}x-1 & y-5 & z-(-7) \\ -3-1 & 6-5 & 3-(-7) \\ -2-1 & 7-5 & 3-(-7)\end{array}\right|=0$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-5 & z+7 \\
-4 & 1 & 10 \\
-3 & 2 & 10
\end{array}\right|=0 \\
& (x-1) \cdot\left|\begin{array}{cc}
1 & 10 \\
2 & 10
\end{array}\right|-(y-5) \cdot\left|\begin{array}{cc}
-4 & 10 \\
-3 & 10
\end{array}\right|+(z+7) \cdot\left|\begin{array}{ll}
-4 & 1 \\
-3 & 2
\end{array}\right|=0 \\
& (x-1) \cdot(-10)-(y-5) \cdot(-10)+(z+7) \cdot(-5)=0 \\
& -10 x+10+10 y-50-5 z-35=0 \\
& -10 x+10 y-5 z-75=0 \\
& 2 x-2 y+z+15=0
\end{aligned}
$$

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:

$$
d=\frac{|2 \cdot 1-2 \cdot(-1)+2+15|}{\sqrt{2^{2}+(-2)^{2}+1^{2}}}=\frac{|2+2+2+15|}{\sqrt{4+4+1}}=\frac{21}{\sqrt{9}}=\frac{21}{3}=7
$$

## Problem Kuznetsov Analytic Geometry 8-31