## Problem Statement

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

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

$M_{2}(3 ; 5 ;-2)$

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

$M_{0}(-3 ; 1 ; 8)$
## Solution

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

$$
\left|\begin{array}{ccc}
x-2 & y-1 & z-4 \\
3-2 & 5-1 & -2-4 \\
-7-2 & -3-1 & 2-4
\end{array}\right|=0
$$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-1 & z-4 \\
1 & 4 & -6 \\
-9 & -4 & -2
\end{array}\right|=0 \\
& (x-2) \cdot\left|\begin{array}{cc}
4 & -6 \\
-4 & -2
\end{array}\right|-(y-1) \cdot\left|\begin{array}{cc}
1 & -6 \\
-9 & -2
\end{array}\right|+(z-4) \cdot\left|\begin{array}{cc}
1 & 4 \\
-9 & -4
\end{array}\right|=0 \\
& (x-2) \cdot(-32)-(y-1) \cdot(-56)+(z-4) \cdot 32=0 \\
& -32 x+64+56 y-56+32 z-128=0 \\
& -32 x+56 y+32 z-120=0 \\
& -4 x+7 y+4 z-15=0
\end{aligned}
$$

The distance $d_{\text {from point }} M_{0}\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {to the plane }} A x+B y+C z+D=0$:

$$
d=\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}
$$

## Find:

$$
d=\frac{|-4 \cdot(-3)+7 \cdot 1+4 \cdot 8-15|}{\sqrt{(-4)^{2}+7^{2}+4^{2}}}=\frac{|12+7+32-15|}{\sqrt{16+49+16}}=\frac{36}{\sqrt{81}}=4
$$

## Problem Kuznetsov Analytic Geometry 8-19