## Problem Statement

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

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

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

$M_{3}(6 ; 3 ; 7)$

$M_{0}(-5 ;-4 ; 8)$
## Solution

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

$\left|\begin{array}{ccc}x-2 & y-3 & z-1 \\ 4-2 & 1-3 & -2-1 \\ 6-2 & 3-3 & 7-1\end{array}\right|=0$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-3 & z-1 \\
2 & -2 & -3 \\
4 & 0 & 6
\end{array}\right|=0 \\
& (x-2) \cdot\left|\begin{array}{cc}
-2 & -3 \\
0 & 6
\end{array}\right|-(y-3) \cdot\left|\begin{array}{cc}
2 & -3 \\
4 & 6
\end{array}\right|+(z-1) \cdot\left|\begin{array}{cc}
2 & -2 \\
4 & 0
\end{array}\right|=0 \\
& (x-2) \cdot(-12)-(y-3) \cdot 24+(z-1) \cdot 8=0 \\
& -12 x+24+(-24) y+72+8 z+(-8)=0 \\
& -12 x-24 y+8 z+88=0 \\
& -3 x-6 y+2 z+22=0
\end{aligned}
$$

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:

$$
d=\frac{|-3 \cdot(-5)-6 \cdot(-4)+2 \cdot 8+22|}{\sqrt{(-3)^{2}+(-6)^{2}+2^{2}}}=\frac{|15+24+16+22|}{\sqrt{9+36+4}}=\frac{77}{\sqrt{49}}=11
$$

## Problem Kuznetsov Analytic Geometry $8-29$