## Problem Statement

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

$$
\begin{aligned}
& M_{1}(14 ; 4 ; 5) \\
& M_{2}(-5 ;-3 ; 2) \\
& M_{3}(-2 ;-6 ;-3) \\
& M_{0}(-1 ;-8 ; 7)
\end{aligned}
$$
## Solution

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

$\left|\begin{array}{ccc}x-14 & y-4 & z-5 \\ -5-14 & -3-4 & 2-5 \\ -2-14 & -6-4 & -3-5\end{array}\right|=0$

Perform transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-14 & y-4 & z-5 \\
-19 & -7 & -3 \\
-16 & -10 & -8
\end{array}\right|=0 \\
& (x-14) \cdot\left|\begin{array}{cc}
-7 & -3 \\
-10 & -8
\end{array}\right|-(y-4) \cdot\left|\begin{array}{cc}
-19 & -3 \\
-16 & -8
\end{array}\right|+(z-5) \cdot\left|\begin{array}{cc}
-19 & -7 \\
-16 & -10
\end{array}\right|=0 \\
& (x-14) \cdot 26-(y-4) \cdot 104+(z-5) \cdot 78=0 \\
& 26 x-364-104 y+416+78 z-390=0 \\
& 26 x-104 y+78 z-338=0 \\
& x-4 y+3 z-13=0
\end{aligned}
$$

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:

$$
d=\frac{|-1-4 \cdot(-8)+3 \cdot 7-13|}{\sqrt{1^{2}+(-4)^{2}+3^{2}}}=\frac{|-1+32+21-13|}{\sqrt{1+16+9}}=\frac{39}{\sqrt{26}}=3 \sqrt{\frac{13}{2}}
$$

## Problem Kuznetsov Analytic Geometry 8-25