## Problem Statement

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

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

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

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

$M_{0}(-3 ;-7 ; 6)$
## Solution

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

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

Perform the transformations:

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

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:

$$
d=\frac{|2 \cdot(-3)+2 \cdot(-7)+(-3) \cdot 6-7|}{\sqrt{2^{2}+2^{2}+(-3)^{2}}}=\frac{|-6-14-18-7|}{\sqrt{4+4+9}}=\frac{45}{\sqrt{17}}
$$

## Problem Kuznetsov Analytic Geometry 8-30

This problem may have a different condition (possibly due to different editions or errors). See below for more details.