## problem statement

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

$$
\begin{aligned}
& M_{1}(3 ; 10 ;-1) \\
& M_{2}(-2 ; 3 ;-5) \\
& M_{3}(-6 ; 0 ;-3) \\
& M_{0}(-6 ; 7 ;-10)
\end{aligned}
$$
## Solution

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

$$
\left|\begin{array}{ccc}
x-3 & y-10 & z-(-1) \\
-2-3 & 3-10 & -5-(-1) \\
-6-3 & 0-10 & -3-(-1)
\end{array}\right|=0
$$

Perform the transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-3 & y-10 & z-(-1) \\
-5 & -7 & -4 \\
-9 & -10 & -2
\end{array}\right|=0 \\
& (x-3) \cdot\left|\begin{array}{cc}
-7 & -4 \\
-10 & -2
\end{array}\right|-(y-10) \cdot\left|\begin{array}{cc}
-5 & -4 \\
-9 & -2
\end{array\right|+(z+1) \cdot\left|\begin{array}{cc}
-5 & -7 \\
-9 & -10
\end{array}\right|=0 \\
& (x-3) \cdot(-26)-(y-10) \cdot(-26)+(z+1) \cdot(-13)=0 \\
& -26 x+78+26 y-260-13 z-13=0 \\
& -26 x+26 y-13 z-195=0 \\
& -2 x+2 y-z-15=0
\end{aligned}
$$

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

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

$$
d=\frac{|-2 \cdot(-6)+2 \cdot 7-\cdot(-10)-15|}{\sqrt{(-2)^{2}+2^{2}+(-1)^{2}}}=\frac{|12+14+10-15|}{\sqrt{4+4+1}}=\frac{21}{\sqrt{9}}=7
$$

## Problem Kuznetsov Analytic Geometry $8-8$