## Problem Statement

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

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

$M_{2}(5 ;-6 ; 0)$

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

$M_{0}(2 ;-10 ; 8)$
## Solution

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

$\left|\begin{array}{ccc}x-2 & y-(-4) & z-(-3) \\ 5-2 & -6-(-4) & 0-(-3) \\ -1-2 & 3-(-4) & -3-(-3)\end{array}\right|=0$

Perform transformations:

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

$$
\begin{aligned}
& (x-2) \cdot(-21)-(y+4) \cdot 9+(z+3) \cdot 15=0 \\
& -21 x+42-9 y-36+15 z+45=0 \\
& -21 x-9 y+15 z+51=0 \\
& -7 x-3 y+5 z+17=0
\end{aligned}
$$

The distance $d$ from a 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:

$$
d=\frac{|-7 \cdot 2-3 \cdot(-10)+5 \cdot 8+17|}{\sqrt{(-7)^{2}+(-3)^{2}+5^{2}}}=\frac{|-14+30+40+17|}{\sqrt{49+9+25}}=\frac{73}{\sqrt{83}}
$$

## Problem Kuznetsov Analytic Geometry 8-14