## Problem Statement

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

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

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

$M_{3}(5; 0; -6)$

$M_{0}(14; -3; 7)$
## Solution

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

$$
\left|\begin{array}{ccc}
x-2 & y-(-1) & z-(-2) \\
1-2 & 2-(-1) & 1-(-2) \\
5-2 & 0-(-1) & -6-(-2)
\end{array}\right|=0
$$

Perform transformations:

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

$$
\begin{aligned}
& (x-2) \cdot\left|\begin{array}{cc}
3 & 3 \\
1 & -4
\end{array}\right|-(y+1) \cdot\left|\begin{array}{cc}
-1 & 3 \\
3 & -4
\end{array}\right|+(z+2) \cdot\left|\begin{array}{cc}
-1 & 3 \\
3 & 1
\end{array}\right|=0 \\
& (x-2) \cdot(-15)-(y+1) \cdot(-5)+(z+2) \cdot(-10)=0 \\
& -15 x+30+5 y+5-10 z-20=0 \\
& -15 x+5 y-10 z+15=0 \\
& -3 x+y-2 z+3=0
\end{aligned}
$$

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:

$$
d=\frac{|-3 \cdot 14-3-2 \cdot 7+3|}{\sqrt{(-3)^{2}+1^{2}+(-2)^{2}}}=\frac{|-42-3-14+3|}{\sqrt{9+1+4}}=\frac{56}{\sqrt{14}}=4 \sqrt{14}
$$

## Problem Kuznetsov Analytic Geometry $8-23$