## Problem Statement

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

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

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

$M_{3}(3; 1; -4)$

$M_{0}(-21; 20; -16)$
## Solution

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

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

Perform the transformations:

$$
\begin{aligned}
& \left|\begin{array}{ccc}
x+2 & y+1 & z+1 \\
2 & 4 & 3 \\
5 & 2 & -3
\end{array}\right|=0 \\
& (x+2) \cdot\left|\begin{array}{cc}
4 & 3 \\
2 & -3
\end{array}\right|-(y+1) \cdot\left|\begin{array}{cc}
2 & 3 \\
5 & -3
\end{array}\right|+(z+1) \cdot\left|\begin{array}{cc}
2 & 4 \\
5 & 2
\end{array}\right|=0 \\
& (x+2) \cdot(-18)-(y+1) \cdot(-21)+(z+1) \cdot(-16)=0 \\
& -18 x-36+21 y+21-16 z-16=0 \\
& -18 x+21 y-16 z-31=0
\end{aligned}
$$

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:

$$
d=\frac{|-18 \cdot(-21)+21 \cdot 20-16 \cdot(-16)-31|}{\sqrt{(-18)^{2}+21^{2}+(-16)^{2}}}=\frac{|378+420+256-31|}{\sqrt{324+441+256}}=\frac{1023}{\sqrt{1021}}
$$

## Problem Kuznetsov Analytic Geometry 8-12