## problem statement

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

$M_{1}(1 ; 3 ; 0)$

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

$M_{3}(3 ; 0 ; 1)$

$M_{0}(4 ; 3 ; 0)$
## Solution

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

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

Perform transformations:

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

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:

$d=\frac{|2 \cdot 4+3-0-5|}{\sqrt{2^{2}+1^{2}+(-1)^{2}}}=\frac{|8+3-5|}{\sqrt{4+1+1}}=\frac{6}{\sqrt{6}}=\sqrt{6}$

## Problem Kuznetsov Analytic Geometry $8-11$