## Problem Statement

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

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

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

$M_{3}(2 ;-1 ; 5)$

$M_{0}(-3 ; 4 ;-5)$
## Solution

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

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

Perform transformations:
$\left|\begin{array}{ccc}x & y+3 & z-1 \\ -4 & 4 & 1 \\ 2 & 2 & 4\end{array}\right|=0$

$x \cdot\left|\begin{array}{ll}4 & 1 \\ 2 & 4\end{array}\right|-(y+3) \cdot\left|\begin{array}{cc}-4 & 1 \\ 2 & 4\end{array}\right|+(z-1) \cdot\left|\begin{array}{cc}-4 & 4 \\ 2 & 2\end{array}\right|=0$

$x \cdot 14-(y+3) \cdot(-18)+(z-1) \cdot(-16)=0$

$14 x+18 y+54-16 z+16=0$

$14 x+18 y-16 z+70=0$

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:

$$
d=\frac{|14 \cdot(-3)+18 \cdot 4-16 \cdot(-5)+70|}{\sqrt{14^{2}+18^{2}+(-16)^{2}}}=\frac{|-42+72+80+70|}{\sqrt{196+324+256}}=\frac{180}{\sqrt{776}}=\frac{90}{\sqrt{194}}
$$

## Problem Kuznetsov Analytic Geometry $8-10$