## problem statement

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

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

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

$M_{3}(2 ;-2 ; 4)$

$M_{0}(2 ; 3 ; 8)$
## Solution

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

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

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-1 & z-2 \\
-2 & 0 & 1 \\
1 & -3 & 2
\end{array}\right|=0 \\
& (x-1) \cdot\left|\begin{array}{cc}
0 & 1 \\
-3 & 2
\end{array}\right|-(y-1) \cdot\left|\begin{array}{cc}
-2 & 1 \\
1 & 2
\end{array}\right|+(z-2) \cdot\left|\begin{array}{cc}
-2 & 0 \\
1 & -3
\end{array}\right|=0 \\
& (x-1) \cdot 3-(y-1) \cdot(-5)+(z-2) \cdot 6=0 \\
& 3 x-3+5 y-5+6 z-12=0 \\
& 3 x+5 y+6 z-20=0
\end{aligned}
$$

The distance $d_{\text {from point }} M_{0}\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {to the plane }} A x+B y+C z+D=0$:

$d=\frac{\left|A x_{0}+B y_{0}+C z_{0}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}$

Find:

$$
d=\frac{|3 \cdot 2+5 \cdot 3+6 \cdot 8-20|}{\sqrt{3^{2}+5^{2}+6^{2}}}=\frac{|6+15+48-20|}{\sqrt{9+25+36}}=\frac{49}{\sqrt{70}}=7 \sqrt{\frac{7}{10}}
$$

## Problem Kuznetsov Analytic Geometry 8-28