## Problem Statement

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

$M_{1}(-4 ; 2 ; 6)$

$M_{2}(2 ;-3 ; 0)$

$M_{3}(-10 ; 5 ; 8)$

$M_{0}(-12 ; 1 ; 8)$
## Solution

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

$$
\left|\begin{array}{ccc}
x-(-4) & y-2 & z-6 \\
2-(-4) & -3-2 & 0-6 \\
-10-(-4) & 5-2 & 8-6
\end{array}\right|=0
$$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+4 & y-2 & z-6 \\
6 & -5 & -6 \\
-6 & 3 & 2
\end{array}\right|=0 \\
& (x+4) \cdot\left|\begin{array}{cc}
-5 & -6 \\
3 & 2
\end{array}\right|-(y-2) \cdot\left|\begin{array}{cc}
6 & -6 \\
-6 & 2
\end{array}\right|+(z-6) \cdot\left|\begin{array}{cc}
6 & -5 \\
-6 & 3
\end{array}\right|=0 \\
& (x+4) \cdot 8-(y-2) \cdot(-24)+(z-6) \cdot(-12)=0 \\
& 8 x+32+24 y-48-12 z+72=0 \\
& 8 x+24 y-12 z+56=0 \\
& 2 x+6 y-3 z+14=0
\end{aligned}
$$

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:

$$
d=\frac{|2 \cdot(-12)+6 \cdot 1-3 \cdot 8+14|}{\sqrt{2^{2}+6^{2}+(-3)^{2}}}=\frac{|-24+6-24+14|}{\sqrt{4+36+9}}=\frac{28}{\sqrt{49}}=4
$$

## Problem Kuznetsov Analytic Geometry $8-17$