## Problem Statement

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

$M_{1}(7 ; 2 ; 4)$

$M_{2}(7 ;-1 ;-2)$

$M_{3}(-5 ;-2 ;-1)$

$M_{0}(10 ; 1 ; 8)$
## Solution

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

$$
\left|\begin{array}{ccc}
x-7 & y-2 & z-4 \\
7-7 & -1-2 & -2-4 \\
-5-7 & -2-2 & -1-4
\end{array}\right|=0
$$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-7 & y-2 & z-4 \\
0 & -3 & -6 \\
-12 & -4 & -5
\end{array}\right|=0 \\
& (x-7) \cdot\left|\begin{array}{ll}
-3 & -6 \\
-4 & -5
\end{array}\right|-(y-2) \cdot\left|\begin{array}{cc}
0 & -6 \\
-12 & -5
\end{array}\right|+(z-4) \cdot\left|\begin{array}{cc}
0 & -3 \\
-12 & -4
\end{array}\right|=0 \\
& (x-7) \cdot(-9)-(y-2) \cdot(-72)+(z-4) \cdot(-36)=0 \\
& -9 x+63+72 y-144-36 z+144=0 \\
& -9 x+72 y-36 z+63=0 \\
& -x+8 y-4 z+7=0
\end{aligned}
$$

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:

$$
d=\frac{|-10+8 \cdot 1-4 \cdot 8+7|}{\sqrt{(-1)^{2}+8^{2}+(-4)^{2}}}=\frac{|-10+8-32+7|}{\sqrt{1+64+16}}=\frac{27}{\sqrt{81}}=3
$$

## Problem Kuznetsov Analytic Geometry 8-18