## Problem Statement

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

$$
\begin{aligned}
& M_{1}(-3 ;-1 ; 1) \\
& M_{2}(-9 ; 1 ;-2) \\
& M_{3}(3 ;-5 ; 4) \\
& M_{0}(-7 ; 0 ;-1)
\end{aligned}
$$
## Solution

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

$$
\left|\begin{array}{ccc}
x-(-3) & y-(-1) & z-1 \\
-9-(-3) & 1-(-1) & -2-1 \\
3-(-3) & -5-(-1) & 4-1
\end{array}\right|=0
$$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+3 & y+1 & z-1 \\
-6 & 2 & -3 \\
6 & -4 & 3
\end{array}\right|=0 \\
& (x+3) \cdot\left|\begin{array}{cc}
2 & -3 \\
-4 & 3
\end{array}\right|-(y+1) \cdot\left|\begin{array}{cc}
-6 & -3 \\
6 & 3
\end{array}\right|+(z-1) \cdot\left|\begin{array}{cc}
-6 & 2 \\
6 & -4
\end{array}\right|=0 \\
& (x+3) \cdot(-6)-(y+1) \cdot 0+(z-1) \cdot 12=0 \\
& -6 x-18+12 z-12=0 \\
& -6 x+12 z-30=0 \\
& -x+2 z-5=0
\end{aligned}
$$

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:

$$
d=\frac{|-1 \cdot(-7)+2 \cdot(-1)-5|}{\sqrt{(-1)^{2}+0^{2}+2^{2}}}=\frac{|7-2-5|}{\sqrt{1+0+4}}=\frac{0}{\sqrt{5}}=0
$$

## Problem Kuznetsov Analytic Geometry $8-3$