## Problem Statement

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

$M_{1}(-3 ;-5 ; 6)$

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

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

$M_{0}(3 ; 6 ; 68)$
## Solution

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

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

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+3 & y+5 & z-6 \\
5 & 6 & -10 \\
3 & 2 & -7
\end{array}\right|=0 \\
& (x+3) \cdot\left|\begin{array}{cc}
6 & -10 \\
2 & -7
\end{array}\right|-(y+5) \cdot\left|\begin{array}{cc}
5 & -10 \\
3 & -7
\end{array}\right|+(z-6) \cdot\left|\begin{array}{cc}
5 & 6 \\
3 & 2
\end{array}\right|=0 \\
& (x+3) \cdot(-22)-(y+5) \cdot(-5)+(z-6) \cdot(-8)=0 \\
& -22 x-66+5 y+25-8 z+48=0 \\
& -22 x+5 y-8 z+7=0
\end{aligned}
$$

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:

$$
d=\frac{|-22 \cdot 3+5 \cdot 6-8 \cdot 68+7|}{\sqrt{(-22)^{2}+5^{2}+(-8)^{2}}}=\frac{|-66+30-544+7|}{\sqrt{484+25+64}}=\frac{573}{\sqrt{573}}=\sqrt{573}
$$

## Problem Kuznetsov Analytic Geometry 8-13