## Problem Statement

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

$M_{1}(-3 ; 4 ;-7)$

$M_{2}(1 ; 5 ;-4)$

$M_{3}(-5 ;-2 ; 0)$

$M_{0}(-12 ; 7 ;-1)$
## Solution

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

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

Perform transformations:

$\left|\begin{array}{ccc}x+3 & y-4 & z+7 \\ 4 & 1 & 3 \\ -2 & -6 & 7\end{array}\right|=0$

$(x+3) \cdot\left|\begin{array}{cc}1 & 3 \\ -6 & 7\end{array}\right|-(y-4) \cdot\left|\begin{array}{cc}4 & 3 \\ -2 & 7\end{array}\right|+(z+7) \cdot\left|\begin{array}{cc}4 & 1 \\ -2 & -6\end{array}\right|=0$

$(x+3) \cdot 25-(y-4) \cdot 34+(z+7) \cdot(-22)=0$

$25 x+75-34 y+136-22 z-154=0$

$25 x-34 y-22 z+57=0$

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:

$$
d=\frac{|25 \cdot(-12)-34 \cdot 7-22 \cdot(-1)+57|}{\sqrt{25^{2}+(-34)^{2}+(-22)^{2}}}=\frac{|-300-238+22+57|}{\sqrt{625+1156+484}}=\frac{459}{\sqrt{2265}}
$$

## Problem Kuznetsov Analytic Geometry 8-1