## Problem Statement

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

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

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

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

$M_{0}(-3 ; 2 ; 7)$
## Solution

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

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

Perform transformations:

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

$$
\begin{aligned}
& (x-1) \cdot\left|\begin{array}{ll}
2 & 0 \\
2 & 2
\end{array}\right|-(y+1) \cdot\left|\begin{array}{ll}
1 & 0 \\
0 & 2
\end{array}\right|+(z-2) \cdot\left|\begin{array}{ll}
1 & 2 \\
0 & 2
\end{array}\right|=0 \\
& (x-1) \cdot 4-(y+1) \cdot 2+(z-2) \cdot 2=0 \\
& 4 x-4-2 y-2+2 z-4=0 \\
& 4 x-2 y+2 z-10=0 \\
& 2 x-y+z-5=0
\end{aligned}
$$

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:

$$
d=\frac{|2 \cdot(-3)-2+7-5|}{\sqrt{2^{2}+(-1)^{2}+1^{2}}}=\frac{|-6-2+7-5|}{\sqrt{4+1+1}}=\frac{6}{\sqrt{6}}=\sqrt{6}
$$

## Problem Kuznetsov Analytic Geometry 8-15