## Problem Statement

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

$M_{1}(1 ; 2 ; 0)$

$M_{2}(3 ; 0 ;-3)$

$M_{3}(5 ; 2 ; 6)$

$M_{0}(-13 ;-8 ; 16)$
## Solution

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

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

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-2 & z \\
2 & -2 & -3 \\
4 & 0 & 6
\end{array}\right|=0 \\
& (x-1) \cdot\left|\begin{array}{cc}
-2 & -3 \\
0 & 6
\end{array}\right|-(y-2) \cdot\left|\begin{array}{cc}
2 & -3 \\
4 & 6
\end{array}\right|+z \cdot\left|\begin{array}{cc}
2 & -2 \\
4 & 0
\end{array}\right|=0 \\
& (x-1) \cdot(-12)-(y-2) \cdot 24+z \cdot 8=0 \\
& -12 x+12-24 y+48+8 z=0 \\
& -12 x-24 y+8 z+60=0 \\
& -3 x-6 y+2 z+15=0
\end{aligned}
$$

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:

$$
d=\frac{|-3 \cdot(-13)-6 \cdot(-8)+2 \cdot 16+15|}{\sqrt{(-3)^{2}+(-6)^{2}+2^{2}}}=\frac{|39+48+32+15|}{\sqrt{9+36+4}}=\frac{134}{7}
$$

## Problem Kuznetsov Analytic Geometry 8-26